Trigonometry — Part 2 MCQs with Explanations

$$\eqalign{ & {\text{Given,}} \cr & n = \frac{{{\text{cos }}\alpha }}{{{\text{sin }}\beta }}{\text{,}}\,\,\,\,\,\,{\text{m}} = \frac{{{\text{cos }}\alpha }}{{{\text{cos }}\beta }} \cr & \Rightarrow {\text{cos }}\alpha = n{\text{ sin }}\beta ,\,and\,\cos \alpha = m\,\cos \beta \cr & \Rightarrow {\text{co}}{{\text{s}}^2}{\text{ }}\alpha = {n^2}{\text{ si}}{{\text{n}}^2}{\text{ }}\beta .....(i) \cr & \Rightarrow {\text{co}}{{\text{s}}^2}\alpha = {m^2}{\text{ co}}{{\text{s}}^2}{\text{ }}\beta .....(ii) \cr & equation{\text{ }}(i) = (ii) \cr & \Rightarrow {n^2}{\text{ si}}{{\text{n}}^2}\beta = {m^2}{\text{ co}}{{\text{s}}^2}\beta \cr & \Rightarrow {n^2}\left( {{\text{1}} - {\text{ co}}{{\text{s}}^2}\beta } \right) = {m^2}{\text{ co}}{{\text{s}}^2}\beta \cr & \Rightarrow {n^2} - {n^2}{\cos ^2}\beta = {m^2}{\cos ^2}\beta \cr & \Rightarrow {n^2} = {m^2}{\cos ^2}\beta + {n^2}{\cos ^2}\beta \cr & \Rightarrow {n^2} = {\cos ^2}\beta \left( {{m^2} + {n^2}} \right) \cr & \Rightarrow {\cos ^2}\beta = \frac{{{n^2}}}{{{m^2} + {n^2}}} \cr} $$

$$\eqalign{ & {\text{Given,}} \cr & {\text{tan}}\theta + \cot \theta = 5 \cr & \Rightarrow {\text{tan}}\theta + \cot \theta = 5 \cr & \Rightarrow {\left( {{\text{tan}}\theta + \cot \theta } \right)^2} = {5^2} \cr & \left( {{\text{Squaring both sides}}} \right) \cr & \Rightarrow {\text{ta}}{{\text{n}}^2}\theta + {\cot ^2}\theta + 2{\text{tan}}\theta \cot \theta = 25 \cr & \Rightarrow {\text{ta}}{{\text{n}}^2}\theta + {\text{co}}{{\text{t}}^2}\theta = {\text{25}} - {\text{2}} \cr & \left[ {\because {\text{tan}}\theta .\cot \theta = 1} \right] \cr & \Rightarrow {\text{ta}}{{\text{n}}^2}\theta + {\text{co}}{{\text{t}}^2}\theta = 23 \cr} $$

$$\eqalign{ & {\text{sin A}} = m{\text{ sin B}} \cr & {\text{si}}{{\text{n}}^2}{\text{A}} = {m^2}{\text{si}}{{\text{n}}^2}{\text{B }}......{\text{(i)}} \cr & {\text{Now, ta}}{{\text{n}}^2}{\text{A}} = {n^2}{\text{ta}}{{\text{n}}^2}{\text{B}} \cr & \frac{{{{\sin }^2}{\text{A}}}}{{{\text{co}}{{\text{s}}^2}{\text{A}}}} = {n^2}\frac{{{{\sin }^2}{\text{B}}}}{{{\text{co}}{{\text{s}}^2}{\text{B}}}} \cr & {\text{from equation (i)}} \cr & \Rightarrow \frac{{1 - {\text{co}}{{\text{s}}^2}{\text{A}}}}{{{{\text{n}}^2}{\text{co}}{{\text{s}}^2}{\text{A}}}} = \frac{{{{\sin }^2}{\text{B}}}}{{{\text{co}}{{\text{s}}^2}{\text{B}}}} \cr & \Rightarrow \frac{{1 - {\text{co}}{{\text{s}}^2}{\text{A}}}}{{{{\text{n}}^2}{\text{co}}{{\text{s}}^2}{\text{A}}}} = \frac{{\frac{{\left( {1 - {\text{co}}{{\text{s}}^2}{\text{A}}} \right)}}{{{m^2}}}}}{{1 - \frac{{{{\sin }^2}{\text{A}}}}{{{m^2}}}}} \cr & \Rightarrow \frac{{1 - {\text{co}}{{\text{s}}^2}{\text{A}}}}{{{{\text{n}}^2}{\text{co}}{{\text{s}}^2}{\text{A}}}} = \frac{{{\text{1}} - {\text{co}}{{\text{s}}^2}{\text{A}}}}{{{m^2} - 1 + {\text{co}}{{\text{s}}^2}{\text{A}}}} \cr & \Rightarrow {m^2} - 1 + {\text{co}}{{\text{s}}^2}{\text{A}} = {{\text{n}}^2}{\text{co}}{{\text{s}}^2}{\text{A}} \cr & \Rightarrow {m^2} - 1 = {\text{co}}{{\text{s}}^2}\theta \left( {{n^2} - 1} \right) \cr & \Rightarrow {\text{co}}{{\text{s}}^2}{\text{A}} = \frac{{{m^2} - 1}}{{{n^2} - 1}} \cr} $$

$$\eqalign{ & {\text{tan }}{{\text{4}}^ \circ }{\text{.tan 4}}{{\text{3}}^ \circ }{\text{.tan 4}}{{\text{7}}^ \circ }{\text{.tan 8}}{{\text{6}}^ \circ } \cr & {\text{Here, }} \cr & {\text{tan 8}}{{\text{6}}^ \circ } = {\text{tan}}\left( {{\text{ 9}}{{\text{0}}^ \circ } - {4^ \circ }} \right) = {\text{cot }}{{\text{4}}^ \circ } \cr & {\text{tan 4}}{{\text{7}}^ \circ } = {\text{tan}}\left( {{\text{ 9}}{{\text{0}}^ \circ } - {{43}^ \circ }} \right) = {\text{cot 4}}{{\text{3}}^ \circ } \cr & {\text{tan }}{{\text{4}}^ \circ }{\text{.cot }}{{\text{4}}^ \circ }{\text{.tan 4}}{{\text{3}}^ \circ }{\text{.cot 4}}{{\text{3}}^ \circ } = 1 \cr} $$

tan 1°.tan 2°.tan 3° .......... tan 89° If A + B = 90° (tan 1°.tan 89°)(tan 2°.tan 88°) .......... (tan 44°.tan 46°).tan 45° 1 × 1 × 1 .......... 1 × 1 = 1

cot10° .cot20° .cot60°.cot70° .cot80° [ In cot A.cot B if A + B = 90° then cot A.cot B = 1 ] (cot10° .cot80°) × (cot20°.cot70°) × cot60° ∴ 1 × 1 × $$\frac{1}{{\sqrt 3 }}$$ = $$\frac{1}{{\sqrt 3 }}$$

$$\cot {18^ \circ }\left( {\cot {{72}^ \circ }.{{\cos }^2}{{22}^ \circ } + \frac{1}{{\tan {{72}^ \circ }.{{\sec }^2}{{68}^ \circ }}}} \right)$$ $$ \Rightarrow \cot {18^ \circ }\left( {\cot {{72}^ \circ }.{{\cos }^2}{{22}^ \circ } + \cot {{72}^ \circ }.{{\cos }^2}{{68}^ \circ }} \right) $$ $$ \Rightarrow \cot {18^ \circ }.\cot {72^ \circ }\left( {{{\cos }^2}{{22}^ \circ } + {{\cos }^2}{{68}^ \circ }} \right) $$ We know that, cotA.cotB = 1 and cos 2 A + cos 2 B = 1( when A + B = 90° ) = 1 × 1 = 1

$$\eqalign{ & {\text{sin}}\left( {2x - {{20}^ \circ }} \right) = {\text{cos}}\left( {2y + {{20}^ \circ }} \right) \cr & \Rightarrow \left( {2x - {{20}^ \circ }} \right) + \left( {2y + {{20}^ \circ }} \right) = {90^ \circ } \cr & \left[ {{\text{If sin A}} = {\text{cos B, then A}} + {\text{B}} = {{90}^ \circ }} \right] \cr & \Rightarrow 2\left( {x + y} \right) = {90^ \circ } \cr & \Rightarrow x + y = {45^ \circ } \cr & \therefore \tan \left( {x + y} \right) \cr & = \tan {45^ \circ } \cr & = 1 \cr} $$

$$\eqalign{ & {\sin ^2}{5^ \circ } + {\sin ^2}{6^ \circ } + \,.....\,{\sin ^2}{84^ \circ } + {\sin ^2}{85^ \circ } \cr & {\text{Number of terms}} \cr & = \frac{{85 - 5}}{1} + 1 \cr & = 80 + 1 \cr & = 81 \cr} $$ $$ = \left( {{{\sin }^2}{5^ \circ } + {{\sin }^2}{{85}^ \circ }} \right)\, + $$ $$\,\left( {{{\sin }^2}{6^ \circ } + {{\sin }^2}{{84}^ \circ }} \right)\, + $$ . . . . . upto 40 pairs $$ + $$ middle term $$\eqalign{ & = {\text{40}} + {\sin ^2}{45^ \circ } \cr & = 40 + \frac{1}{2} \cr & = 40\frac{1}{2} \cr} $$ Alternate: In case, when series is in the form of sin 2 θ or cos 2 θ, then sum of series will always be half of number of terms. $$ = \frac{{81}}{2} = 40\frac{1}{2}$$

Given, (sin 2 5° + sin 2 10° + sin 2 15° + . . . . . . + sin 2 85° ) + sin 2 90° We know that sinθ = cos(90° - θ ) Therefore sin 2 85° = cos 2 (90° - 85°) = cos 2 5° Similary sin 2 60° = cos 2 (90° - 60°) = cos 2 40° And we also know that sin 2 θ + cos 2 θ = 1 There are 8 pair in given equation sin 2 5° + cos 2 5° = 1 sin 2 10° + cos 2 10° = 1 . . . . . . . . . . . . . . . . . . . . . . . . sin 2 40° + cos 2 40° = 1 i.e. 8 + sin 2 45° + sin 2 90° ⇒ 8 + $$\frac{1}{2}$$ + 1 = $$9\frac{1}{2}$$ Short Trick: $$\left( {si{n^2}{5^ \circ }\, + \,{{\sin }^2}{{10}^ \circ }\, + \,{{\sin }^2}{{15}^ \circ }\, + \,.....\,{{\sin }^2}{{85}^ \circ }} \right)\, + $$ $$\,{\sin ^2}90$$ $$\eqalign{ & {\text{Number of terms}} \cr & = \left\{ {\left( {\frac{{85 - 5}}{5}} \right) + 1} \right\}{\text{ + 1}} \cr & {\text{ = }}\frac{{17}}{2}{\text{ + 1}} \cr & {\text{Sum of series}} = 9\frac{1}{2} \cr} $$

$$\eqalign{ & {\text{sin }}\alpha .\sec \left( {{{30}^ \circ } + \alpha } \right) = 1 \cr & {\bf{Shortcut\,\, method:}} \cr & {\text{Put }}\alpha {\text{ value between }}{{\text{0}}^ \circ }{\text{ to 6}}{{\text{0}}^ \circ } \cr & {\text{If }}\alpha = {30^ \circ } \cr & \Rightarrow {\text{sin 3}}{0^ \circ }.\sec \left( {{{30}^ \circ } + {{30}^ \circ }} \right) = 1 \cr & \Rightarrow {\text{sin 3}}{0^ \circ }.\sec {60^ \circ } = 1 \cr & \Rightarrow \frac{1}{2} \times 2 \cr & \Rightarrow 1 \cr & 1 = 1\left( {{\text{Satisfy}}} \right) \cr & {\text{So}},\alpha = {30^ \circ } \cr & = \sin \alpha + \cos 2\alpha \cr & = \sin {30^ \circ } + \cos 2 \times {30^ \circ } \cr & = \sin {30^ \circ } + \cos {60^ \circ } \cr & = \frac{1}{2} + \frac{1}{2} \cr & = 1 \cr } $$

$$\eqalign{ & {\text{A + B}} = {90^ \circ } \cr & {\text{B}} = {90^ \circ } - {\text{A}} \cr & {\sec ^2}A + se{c^2}{\text{B}} - {\text{se}}{{\text{c}}^2}{\text{A}}{\text{.se}}{{\text{c}}^2}{\text{B}} \cr} $$ $$ = {\sec ^2}A + se{c^2}\left( {{{90}^ \circ } - {\text{A}}} \right) - $$ $${\text{se}}{{\text{c}}^2}{\text{A}}{\text{.se}}{{\text{c}}^2}$$ $$\left( {{{90}^ \circ } - {\text{A}}} \right)$$ $$\eqalign{ & = {\sec ^2}A + {\operatorname{cosec} ^2}{\text{A}} - {\text{se}}{{\text{c}}^2}{\text{A}}{\text{.cose}}{{\text{c}}^2}{\text{A}} \cr & = \frac{1}{{{{\cos }^2}{\text{A}}}} + \frac{1}{{si{n^2}{\text{A}}}} - \frac{1}{{{\text{co}}{{\text{s}}^2}{\text{A}}}} \times \frac{1}{{{\text{si}}{{\text{n}}^2}{\text{A}}}} \cr & = \frac{{{{\sin }^2}{\text{A + co}}{{\text{s}}^2}{\text{A}}}}{{{\text{co}}{{\text{s}}^2}{\text{A}}{\text{.}}{{\sin }^2}{\text{A}}}} - \frac{1}{{{\text{co}}{{\text{s}}^2}{\text{A}}{\text{.}}{{\sin }^2}{\text{A}}}} \cr & = \frac{1}{{{\text{co}}{{\text{s}}^2}{\text{A}}.{{\sin }^2}{\text{A}}}} - \frac{1}{{{\text{co}}{{\text{s}}^2}{\text{A}}.{{\sin }^2}{\text{A}}}} \cr & = 0 \cr} $$

cotθ. tan(90° - θ) - sec(90° - θ)cosecθ + (sin 2 25° + sin 2 65°) + $$\sqrt 3 $$ (tan5°. tan15°. tan30°. tan75°. tan85°) = cotθ. cotθ - cosecθ. cosecθ + (sin 2 25° + cos 2 25° ) + $$\sqrt 3 $$ [(tan5°. tan85°) . (tan15°. tan75°). tan30°] $$ = \left( {{\text{co}}{{\text{t}}^2}\theta - {\text{cose}}{{\text{c}}^2}\theta } \right) + \left( 1 \right) + \sqrt 3 \left( {1.1.\frac{1}{{\sqrt 3 }}} \right)$$ $$\left[ {{\text{tan A}}{\text{.tan B}} = {\text{1, If A}} + {\text{B}} = {{90}^ \circ }} \right]$$ $$\eqalign{ & = \left( { - 1} \right) + \left( 1 \right) + \sqrt 3 \times \frac{1}{{\sqrt 3 }} \cr & = - 1 + 1 + 1 \cr & = 1 \cr} $$

$$\eqalign{ & {\text{cos}}\theta .{\text{cosec2}}{{\text{3}}^ \circ } = {\text{1}} \cr & {\text{If cosA}}.{\text{cosecB}} = {\text{1}} \cr & {\text{Then, A + B}} = {90^ \circ } \cr & {\text{So,}} \cr & \theta {\text{ + 2}}{{\text{3}}^ \circ } = {90^ \circ } \cr & \theta = {67^ \circ } \cr} $$

$$\eqalign{ & {\text{A}} + {\text{B}} + {\text{C}} = \pi = {\text{18}}{0^ \circ }{\text{ }} \cr & \Rightarrow \frac{{{\text{A + B}}}}{2} = \frac{{{{180}^ \circ }}}{2} - \frac{{\text{C}}}{2} \cr & \Rightarrow {\text{sin }}\left( {\frac{{{\text{A + B}}}}{2}} \right) \cr & \Rightarrow {\text{sin }}\left( {\frac{\pi }{2} - \frac{{\text{C}}}{2}} \right) \cr & \Rightarrow {\text{cos}}\frac{{\text{C}}}{2} \cr & \cr & {\bf{Similarly:}} \cr & {\text{cos }}\left( {\frac{{{\text{A}} + {\text{B}}}}{2}} \right) = {\text{sin}}\frac{{\text{C}}}{2} \cr & {\text{cot }}\left( {\frac{{{\text{A}} + {\text{B}}}}{2}} \right) = \tan \frac{{\text{C}}}{2} \cr & {\text{tan }}\left( {\frac{{{\text{A}} + {\text{B}}}}{2}} \right) = \cot \frac{{\text{C}}}{2} \cr & {\text{So, option C is incorrect}}{\text{.}} \cr} $$

$$\eqalign{ & \tan 2\theta .\tan 3\theta = 1 \cr & \left( {2\theta + 3\theta } \right) = {90^ \circ } \cr & 5\theta = {90^ \circ } \cr & \left[ {{\text{If tan A}}{\text{.tan B}} = {\text{1}}} \right] \cr & \left[ {{\text{then, A}} + {\text{B}} = {{90}^ \circ }} \right] \cr & \Rightarrow \left( {{\text{2co}}{{\text{s}}^2}\frac{{5\theta }}{2} - 1} \right) \cr & \Rightarrow {\text{2co}}{{\text{s}}^2}\frac{{{{90}^ \circ }}}{2} - 1 \cr & \Rightarrow {\text{2co}}{{\text{s}}^2}{45^ \circ } - 1 \cr & \Rightarrow \frac{{\text{2}}}{2} - 1 \cr & \Rightarrow 0 \cr} $$

$$\eqalign{ & \tan 7\theta .\tan 2\theta = 1 \cr & \left[ {{\text{If tan A}}{\text{.tan B}} = {\text{1}}} \right] \cr & ({\text{then, A}} + {\text{B}} = {90^ \circ }) \cr & \left( {7\theta + 2\theta } \right) = {90^ \circ } \cr & 9\theta = {90^ \circ } \cr & \theta = {10^ \circ } \cr & \Rightarrow \tan 3\theta \cr & \Rightarrow \tan {30^ \circ } \cr & \Rightarrow \frac{1}{{\sqrt 3 }} \cr} $$

$$\eqalign{ & {\text{sin}}\left( {{{60}^ \circ } - \theta } \right) = {\text{cos}}\left( {\psi - {{30}^ \circ }} \right) \cr & \Rightarrow \left( {{{60}^ \circ } - \theta } \right) + \left( {\psi - {{30}^ \circ }} \right) = {90^ \circ } \cr & \left[ {{\text{If sin A}} = {\text{cos B}}\,{\text{then, A}} + {\text{B}} = {{90}^ \circ }} \right] \cr & \Rightarrow \left( {\psi - \theta } \right) = {90^ \circ } - {30^ \circ } \cr & \Rightarrow \left( {\psi - \theta } \right) = {60^ \circ } \cr & \Rightarrow \tan \left( {\psi - \theta } \right) = \tan {60^ \circ } \cr & \Rightarrow \tan {60^ \circ } = \sqrt 3 \cr} $$

3cos80°.cosec10° + 2cos59°.cosec31° [If A + B = 90° then, cosA.cosecB = 1] (3 × 1 + 2 × 1 = 5) (the value of sinθ is always between -1 to +1) $$ \Rightarrow {\text{3cos8}}{{\text{0}}^ \circ }\frac{1}{{\sin {{10}^ \circ }}}{\text{ + 2cos5}}{{\text{9}}^ \circ }\frac{1}{{{\text{sin 3}}{{\text{1}}^ \circ }}}$$ $$ \Rightarrow {\text{3cos8}}{{\text{0}}^ \circ }\frac{1}{{\sin \left( {{{90}^ \circ } - {{80}^ \circ }} \right)}}{\text{ + }}$$ $${\text{2cos5}}{{\text{9}}^ \circ }$$ $$\frac{1}{{{\text{sin }}{{\left( {{{90}^ \circ } - {{59}^ \circ }} \right)}^ \circ }}}$$ $$\eqalign{ & \Rightarrow 3 + 2 \cr & \Rightarrow 5 \cr} $$

$$\frac{{2{\text{sin }}{{68}^ \circ }}}{{{\text{cos 2}}{{\text{2}}^ \circ }}}$$ $$ - $$ $$\frac{{2{\text{cot 1}}{5^ \circ }}}{{5\tan {{75}^ \circ }}}$$ $$ - $$ $$\frac{{3\tan {{45}^ \circ }.\tan {{20}^ \circ }.\tan {{40}^ \circ }.\tan {{50}^ \circ }.\tan {{70}^ \circ }}}{5}$$ $$ \Rightarrow \frac{{2{\text{sin }}{{68}^ \circ }}}{{{\text{cos }}\left( {{{90}^ \circ } - {{68}^ \circ }} \right)}} - $$ $$\frac{{2{\text{cot 1}}{5^ \circ }}}{{5\tan \left( {{{90}^ \circ } - {{15}^ \circ }} \right)}} - $$ $$\frac{{3.1\left( {\tan {{20}^ \circ }.\tan {{70}^ \circ }} \right).\left( {\tan {{40}^ \circ }.\tan {{50}^ \circ }} \right)}}{5}$$ $$\eqalign{ & \Rightarrow \frac{{2{\text{sin }}{{68}^ \circ }}}{{{\text{sin 6}}{{\text{8}}^ \circ }}} - \frac{{2{\text{cot 1}}{5^ \circ }}}{{5\cot {{15}^ \circ }}} - \frac{{3 \times 1 \times 1 \times 1}}{5} \cr & \left[ {{\text{If tan A}}{\text{.tan B}} = {\text{1 then, A}} + {\text{B}} = {{90}^ \circ }} \right] \cr & \Rightarrow 2 - \frac{2}{5} - \frac{3}{5} \cr & \Rightarrow 1 \cr} $$

$$\eqalign{ & \tan {10^ \circ }.\tan {15^ \circ }.\tan {75^ \circ }.\tan {80^ \circ } \cr & \Rightarrow \left( {\tan {{10}^ \circ }.\tan {{80}^ \circ }} \right).\left( {\tan {{15}^ \circ }.\tan {{75}^ \circ }} \right) \cr & \Rightarrow 1 \times 1 \cr & \left[ {{\text{If tan A}}{\text{.tan B}} = {\text{1}}{\text{. then, A}} + {\text{B}} = {{90}^ \circ }} \right] \cr & \Rightarrow 1 \cr} $$

$$\eqalign{ & \Rightarrow \sin {\text{ }}7x = \cos 11x \cr & \Rightarrow 7x + 11x = {90^ \circ } \cr & \Rightarrow 18x = {90^ \circ } \cr & \Rightarrow x = {5^ \circ } \cr & \Rightarrow \tan 9x + \cot 9x \cr & \Rightarrow \tan {45^ \circ } + \cot {45^ \circ } \cr & \Rightarrow 1 + 1 \cr & \Rightarrow 2 \cr} $$

$$\eqalign{ & {\text{If A + B}} = {90^ \circ } \cr & {\text{Then,}}{\sin ^2}{\text{A + }}{\sin ^2}{\text{B}} = 1 \cr & \Rightarrow 7{\frac{1}{2}^ \circ } + 82{\frac{1}{2}^ \circ } \cr & \Rightarrow {90^ \circ } \cr & \Rightarrow 1\left[ {{\text{sin }}{{90}^ \circ } = 1} \right] \cr} $$

$$\eqalign{ & \Rightarrow {\sin ^2}{65^ \circ } + {\sin ^2}{25^ \circ } + {\cos ^2}{35^ \circ } + {\cos ^2}{55^ \circ } \cr & \Rightarrow {\sin ^2}{65^ \circ } + {\sin ^2}\left( {{{90}^ \circ } - {{65}^ \circ }} \right) + \left[ {{{\cos }^2}{{35}^ \circ } + {{\cos }^2}\left( {{{90}^ \circ } - {{35}^ \circ }} \right)} \right] \cr & \Rightarrow \left( {{{\sin }^2}{{65}^ \circ } + {{\cos }^2}{{65}^ \circ }} \right) + \left( {{{\cos }^2}{{35}^ \circ } + si{n^2}{{35}^ \circ }} \right) \cr & \Rightarrow 1 + 1 \cr & \Rightarrow 2 \cr} $$

= (tan 2 α + 1 )sin 2 β = (tan 2 45° + 1 )sin 2 45° $$ = (1 + 1) {\left( {\frac{1}{{\sqrt 2 }}} \right)^2}$$ $$ = 2 \times \frac{1}{2} = 1$$

$$\eqalign{ & \Rightarrow {\text{sin 3A}} = {\text{cos}}\left( {{\text{A}} - {{26}^ \circ }} \right) \cr & \Rightarrow {\text{3A + A}} - {\text{2}}{{\text{6}}^ \circ }{\text{ = }}{90^ \circ } \cr & \left[ {{\text{If sin A}} = {\text{cos B}}{\text{. then, A}} + {\text{B}} = {{90}^ \circ }} \right] \cr & \Rightarrow 4{\text{A = 11}}{6^ \circ } \cr & \Rightarrow {\text{A}} = {29^ \circ } \cr} $$

$$\eqalign{ & {\text{sin}}\left( {\theta + {{18}^ \circ }} \right) = \cos {60^ \circ }\left( {{0^ \circ } < \theta < {{90}^ \circ }} \right) \cr & \Rightarrow \theta + {18^ \circ } + {60^ \circ } = {90^ \circ } \cr & \Rightarrow \theta = {12^ \circ } \cr & \Rightarrow {\text{cos 5}}\theta {\text{ }} \cr & \Rightarrow {\text{cos }}{60^ \circ } \cr & \Rightarrow \frac{1}{2} \cr} $$

$$\eqalign{ & {\text{We know that }} \cr & {\text{tan}}\left( {{{90}^ \circ } - \theta } \right) = {\text{cot}}\theta \cr & {\text{and, cot}}\left( {{{90}^ \circ } - \theta } \right) = {\text{tan}}\theta \cr & \Rightarrow {\text{tan}}\left( {4\theta - {{50}^ \circ }} \right) = {\text{cot}}\left( {{{50}^ \circ } - \theta } \right) \cr & \Rightarrow \cot \left[ {{{90}^ \circ } - \left( {4\theta - {{50}^ \circ }} \right)} \right] = {\text{cot}}\left( {{{50}^ \circ } - \theta } \right) \cr & \Rightarrow {90^ \circ } - \left( {4\theta - {{50}^ \circ }} \right) = \left( {{{50}^ \circ } - \theta } \right) \cr & \Rightarrow {90^ \circ } - 4\theta + {50^ \circ } = {50^ \circ } - \theta \cr & \Rightarrow {90^ \circ } = 3\theta \cr & {\text{then}},\theta = {30^ \circ } \cr} $$

$$\frac{{{{\left( {\tan {{20}^ \circ }} \right)}^2}}}{{{{\left( {{\text{cosec 7}}{0^ \circ }} \right)}^2}}}$$ $$ + $$ $$\frac{{{{\left( {\cot {{20}^ \circ }} \right)}^2}}}{{{{\left( {{\text{sec 7}}{0^ \circ }} \right)}^2}}}$$ $$ + $$ $$2\tan {15^ \circ }$$ . $$\tan {45^ \circ }$$ . $$\tan {75^ \circ }$$ $$\eqalign{ & \Rightarrow \frac{{{{\left( {\tan {{20}^ \circ }} \right)}^2}}}{{{\text{se}}{{\text{c}}^2}{{20}^ \circ }}} + \frac{{{{\left( {\cot {{20}^ \circ }} \right)}^2}}}{{{\text{cose}}{{\text{c}}^2}{{20}^ \circ }}} + 2\tan {15^ \circ }.\tan {75^ \circ } \cr & \left[ {{\text{tan 1}}{5^ \circ }{\text{.tan 7}}{5^ \circ } = {\text{1}}{\text{. If, A}} + {\text{B}} = {{90}^ \circ }} \right] \cr & \Rightarrow \left( {{{\sin }^2}{{20}^ \circ } + {\text{co}}{{\text{s}}^2}{{20}^ \circ }} \right) + 2 \cr & \Rightarrow 1 + 2 \cr & \Rightarrow 3 \cr} $$

$$\eqalign{ & {\left( {\frac{{{\text{sin 4}}{{\text{7}}^ \circ }}}{{\cos {{43}^ \circ }}}} \right)^2} + {\left( {\frac{{\cos {{43}^ \circ }}}{{{\text{sin }}{{47}^ \circ }}}} \right)^2} - 4{\text{co}}{{\text{s}}^2}{45^ \circ } \cr & = {\left( {\frac{{{\text{cos 4}}{{\text{3}}^ \circ }}}{{\cos {{43}^ \circ }}}} \right)^2} + {\left( {\frac{{sin{{47}^ \circ }}}{{{\text{sin }}{{47}^ \circ }}}} \right)^2} - 4 \times \frac{1}{2} \cr & = 1 + 1 - 2 \cr & = 0 \cr & {\bf{Note:}} \cr & \left( {\sin \left( {{{90}^ \circ } - \theta } \right)} \right) = \cos \theta \cr} $$

sin 2 1° + sin 2 11° + sin 2 21° + sin 2 31° + sin 2 41° + sin 2 45° + sin 2 49° + sin 2 59° + sin 2 69° + sin 2 79° + sin 2 89° = (sin 2 1° + sin 2 89°) + (sin 2 11° + sin 2 79°) + (sin 2 21° + sin 2 69°) + (sin 2 31° + sin 2 59°) + (sin 2 41° + sin 2 49°) + sin 2 45° = 1 + 1 + 1 + 1 + 1 + $$\frac{1}{2}$$ [sin 2 A + sin 2 B = 1. If, A + B = 90°] = $$5\frac{1}{2}$$

$$\eqalign{ & \cos {20^ \circ } = m{\text{ }} \cr & \cos {70^ \circ } = n \cr & {\text{So,}} \cr & \Leftrightarrow {m^2} + {n^2} = {\text{co}}{{\text{s}}^2}{20^ \circ } + {\text{co}}{{\text{s}}^2}{70^ \circ } \cr & \left[ {{\text{If co}}{{\text{s}}^2}{\text{A + co}}{{\text{s}}^2}{\text{B}} = {\text{1}}} \right] \cr & ({\text{If, A}} + {\text{B}} = {90^ \circ }) \cr & \Leftrightarrow 1 \cr} $$

$$\eqalign{ & {\text{sin}}\left( {{{90}^ \circ } - \theta } \right) + {\text{cos}}\theta = \sqrt 2 {\text{cos}}\left( {{{90}^ \circ } - \theta } \right) \cr & \Rightarrow {\text{cos}}\theta + {\text{cos}}\theta = \sqrt 2 \sin \theta \cr & \Rightarrow \frac{{2\cos \theta }}{{\sin \theta }} = \sqrt 2 \cr & \Rightarrow \cot \theta = \frac{{1 \to {\text{B}}}}{{\sqrt 2 \to {\text{P}}}} \cr & {\text{So, H}} \to \text{alignment} \cr & \therefore {\text{cosec}}\theta = \frac{{\text{H}}}{{\text{P}}} = \frac{{\sqrt 3 }}{{\sqrt 2 }} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \sqrt {\frac{3}{2}} \cr} $$

$$\eqalign{ & {\text{sin A}} - \cos {\text{A}} = \frac{{\sqrt 3 - 1}}{2} \cr & {\bf{Shortcut \,\,method:}} \cr & {\text{Put, }}\theta = {60^ \circ } \cr & \Rightarrow {\text{sin A}} - \cos {\text{A}} = \frac{{\sqrt 3 - 1}}{2} \cr & \Rightarrow {\text{sin }}{60^ \circ } - \cos {60^ \circ } = \frac{{\sqrt 3 - 1}}{2} \cr & \Rightarrow \frac{{\sqrt 3 }}{2} - \frac{1}{2} = \frac{{\sqrt 3 - 1}}{2} \cr & \Rightarrow \frac{{\sqrt 3 - 1}}{2} = \frac{{\sqrt 3 - 1}}{2}({\text{Matched}}) \cr & Hence, \cr & {\text{sin A}}.cos{\text{A}} \cr & \Rightarrow \frac{{\sqrt 3 }}{2} \times \frac{1}{2} \cr & \Rightarrow \frac{{\sqrt 3 }}{4} \cr & \cr & {\bf{Alternate:}} \cr & {\text{sin A}} - \cos {\text{A}} = \frac{{\sqrt 3 - 1}}{2} \cr & {\text{Squaring both side,}} \cr & \Rightarrow {\text{ si}}{{\text{n}}^2}{\text{ A + }}{\cos ^2}{\text{A}} - {\text{2}}{\text{.sin A}}.cos{\text{A}} = {\left( {\frac{{\sqrt 3 - 1}}{2}} \right)^2} \cr & \Rightarrow 1 - 2{\text{sin A}}.cos{\text{A = }}\frac{{3 + 1 - 2\sqrt 3 }}{4} \cr & \Rightarrow 2{\text{sin A}}.cos{\text{A}} = 1 - 2\frac{{\left( {2 - \sqrt 3 } \right)}}{4} \cr & \Rightarrow 2{\text{sin A}}.cos{\text{A}} = \frac{{2 - 2 + \sqrt 3 }}{2} \cr & \Rightarrow {\text{sin A}}.cos{\text{A}} = \frac{{\sqrt 3 }}{4} \cr} $$

$$\eqalign{ & \frac{{{{\sec }^2}{{70}^ \circ } - {\text{co}}{{\text{t}}^2}{{20}^ \circ }}}{{2\left( {{\text{cose}}{{\text{c}}^2}{{59}^ \circ } - {{\tan }^2}{{31}^ \circ }} \right)}} = \frac{2}{m} \cr & \Rightarrow \frac{{{{\sec }^2}{{70}^ \circ } - {\text{co}}{{\text{t}}^2}\left( {{{90}^ \circ } - {{70}^ \circ }} \right)}}{{2\left( {{\text{cose}}{{\text{c}}^2}{{59}^ \circ } - {{\tan }^2}\left( {{{90}^ \circ } - {{59}^ \circ }} \right)} \right)}} = \frac{2}{m} \cr & \Rightarrow \frac{{{{\sec }^2}{{70}^ \circ } - {\text{ta}}{{\text{n}}^2}{{70}^ \circ }}}{{2\left( {{\text{cose}}{{\text{c}}^2}{{59}^ \circ } - {{\cot }^2}{{59}^ \circ }} \right)}} = \frac{2}{m} \cr & \Rightarrow \frac{1}{2} = \frac{2}{m}\left[ {{{\sec }^2}\theta - {\text{ta}}{{\text{n}}^2}\theta = 1} \right] \cr & (cose{c^2}\theta - {\cot ^2}\theta = 1) \cr & \Rightarrow m = 2 \times 2 \cr & \Rightarrow m = 4 \cr} $$

$$\eqalign{ & \frac{{{\text{tan}}\theta + \cot \theta }}{{{\text{tan}}\theta - \cot \theta }} = 2 \cr & {\text{By componendo and dividendo}} \cr & \Rightarrow \frac{{2{\text{tan}}\theta }}{{2{\text{cos}}\theta }} = \frac{3}{1} \cr & \Rightarrow \frac{{\sin \theta }}{{{\text{cos}}\theta }} \times \frac{{\sin \theta }}{{{\text{cos}}\theta }} = 3 \cr & \Rightarrow {\sin ^2}\theta = 3{\text{co}}{{\text{s}}^2}\theta \cr & \Rightarrow {\sin ^2}\theta = 3\left( {1 - {{\sin }^2}\theta } \right) \cr & \Rightarrow 4{\sin ^2}\theta = 3 \cr & \Rightarrow {\sin ^2}\theta \Rightarrow \frac{3}{4} \cr & \Rightarrow {\text{sin }}\theta = \frac{{\sqrt 3 }}{2} \cr & \cr & {\bf{Alternate:}} \cr & \Rightarrow \frac{{{\text{tan}}\theta + \cot \theta }}{{{\text{tan}}\theta - \cot \theta }} = 2 \cr & {\text{By C and D}} \cr & \Rightarrow \frac{{{\text{tan}}\theta }}{{\cot \theta }} = \frac{3}{1} \cr & \Rightarrow {\text{ta}}{{\text{n}}^2}\theta = 3 \cr & \Rightarrow {\text{tan}}\theta = \sqrt 3 \cr & \theta = {60^ \circ } \cr & \Rightarrow \sin \theta \cr & \Rightarrow {\text{sin }}{60^ \circ } \cr & \Rightarrow \frac{{\sqrt 3 }}{2} \cr} $$

If in the any question componendo and dividendo already used as $$\frac{{a + b}}{{a - b}} = \frac{m}{n}$$ If second time you also want to apply componendo dividendo rule then result will be $$\eqalign{ & \frac{a}{b} = \frac{{m + n}}{{m - n}} \cr & \Leftrightarrow \frac{{\sin \theta + \cos \theta }}{{{\text{sin}}\theta - \cos \theta }} = 3 \cr & \Leftrightarrow \sin \theta + {\text{cos}}\theta = 3\sin \theta - 3{\text{cos}}\theta \cr & \Leftrightarrow 2\sin \theta = 4{\text{cos}}\theta \cr & \Leftrightarrow \frac{{\sin \theta }}{{{\text{cos}}\theta }} = \frac{2}{1} \cr & \Leftrightarrow {\text{tan}}\theta = \frac{2}{1} = \frac{{\text{P}}}{{\text{B}}} \cr & \left[ {\therefore {\text{sin}}\theta = \frac{{\text{P}}}{{\text{H}}} = \frac{2}{{\sqrt 5 }}{\text{ and cos}}\theta = \frac{{\text{B}}}{{\text{H}}} = \frac{1}{{\sqrt 5 }}} \right] \cr} $$ $$\eqalign{ & \Leftrightarrow {\sin ^4}\theta - {\text{co}}{{\text{s}}^4}\theta \cr & \Leftrightarrow \left( {{{\sin }^2}\theta + {\text{co}}{{\text{s}}^2}\theta } \right)\left( {{{\sin }^2}\theta - {\text{co}}{{\text{s}}^2}\theta } \right) \cr & \Leftrightarrow 1\left( {{{\sin }^2}\theta - {\text{co}}{{\text{s}}^2}\theta } \right) \cr & \Leftrightarrow {\left( {\frac{2}{{\sqrt 5 }}} \right)^2} - {\left( {\frac{1}{{\sqrt 5 }}} \right)^2} \cr & \Leftrightarrow \frac{4}{5} - \frac{1}{5} \cr & \Leftrightarrow \frac{3}{5} \cr} $$

$${\text{ tan 1}}{5^ \circ }{\text{cot 7}}{5^ \circ } + {\text{tan 7}}{5^ \circ }{\text{cot 1}}{5^ \circ }$$ $$ = {\text{ tan 1}}{5^ \circ }{\text{cot }}\left( {{{90}^ \circ } - {{15}^ \circ }} \right) + $$ $${\text{tan}}{\left( {{{90}^ \circ } - 15} \right)^ \circ }$$ $${\text{cot1}}{5^ \circ }$$ $$\eqalign{ & = {\text{ ta}}{{\text{n}}^2}{\text{1}}{5^ \circ } + {\text{co}}{{\text{t}}^2}{\text{1}}{5^ \circ } \cr & = {\text{ta}}{{\text{n}}^2}{15^ \circ } + {\text{co}}{{\text{t}}^2}{15^ \circ }\,.....(i) \cr & \left[ {{\bf{Formula}}} \right] \cr & \cot \left( {{{90}^ \circ } - \theta } \right) = \tan \theta \cr & \tan \left( {{{90}^ \circ } - \theta } \right) = \cot \theta \cr & {\text{Put value of tan1}}{5^ \circ } \cr & \cot {15^ \circ } = \frac{1}{{{\text{tan1}}{5^ \circ }}} \cr & \cot {15^ \circ } = \frac{1}{{\left( {2 - \sqrt 3 } \right)}} \cr & \cot {15^ \circ } = \frac{1}{{\left( {2 - \sqrt 3 } \right)}} \times \frac{{\left( {2 + \sqrt 3 } \right)}}{{\left( {2 + \sqrt 3 } \right)}} \cr & \cot {15^ \circ } = 2 + \sqrt 3 \cr & {\text{Now put value in equation (i)}} \cr & {\text{ tan 1}}{5^ \circ } + {\text{cot 1}}{5^ \circ } \cr & = {\left( {2 - \sqrt 3 } \right)^2} + {\left( {2 + \sqrt 3 } \right)^2} \cr & = 4 + 3 - 4\sqrt 3 + 4 + 3 + 4\sqrt 3 \cr & = 14 \cr} $$

$$\eqalign{ & \Leftrightarrow \frac{{\text{A}}}{{\text{B}}} = \frac{{{\text{tan1}}{{\text{1}}^ \circ }{\text{.tan2}}{{\text{9}}^ \circ }}}{{{\text{2cot}}{{61}^ \circ }.\cot {{79}^ \circ }}} \cr & \Leftrightarrow \frac{{\text{A}}}{{\text{B}}} = \frac{{{\text{tan1}}{{\text{1}}^ \circ }{\text{.tan2}}{{\text{9}}^ \circ }}}{{{\text{2}}\left[ {{\text{cot}}\left( {{{90}^ \circ } - {{29}^ \circ }} \right).\cot \left( {{{90}^ \circ } - {{11}^ \circ }} \right)} \right]}} \cr & \Leftrightarrow \frac{{\text{A}}}{{\text{B}}} = \frac{{{\text{tan1}}{{\text{1}}^ \circ }{\text{.tan2}}{{\text{9}}^ \circ }}}{{{\text{2tan1}}{{\text{1}}^ \circ }.tan{{29}^ \circ }}} \cr & \Leftrightarrow \frac{{\text{A}}}{{\text{B}}} = \frac{1}{2} \cr & \Leftrightarrow 2{\text{A}} = {\text{B}} \cr} $$

$$\eqalign{ & {\text{ tan}}\left( {\frac{\pi }{2} - \frac{\theta }{2}} \right) = \sqrt 3 \cr & \Rightarrow {\text{tan}}\left( {{{90}^ \circ } - \frac{\theta }{2}} \right) = \sqrt 3 \,\,\left[ {\pi = {{180}^ \circ }} \right] \cr & \Rightarrow \cot \frac{\theta }{2} = \sqrt 3 \cr & \Rightarrow \cot \frac{\theta }{2} = \cot {30^ \circ } \cr & \Rightarrow \frac{\theta }{2} = {30^ \circ } \cr & \Rightarrow \theta = {60^ \circ } \cr & \Rightarrow \cos {60^ \circ } = \frac{1}{2} \cr} $$