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

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)
  $$ \Rightarrow {\text{3cos8}}{{\text{0}}^ \circ }\frac{1}{{\sin {{10}^ \circ }}}{\text{ + 2cos5}}{{\text{9}}^ \circ }\frac{1}{{{\text{sin 3}}{{\text{1}}^ \circ }}}$$
  $$ \Rightarrow {\text{3cos8}}{{\text{0}}^ \circ }\frac{1}{{\sin \left( {{{90}^ \circ } - {{80}^ \circ }} \right)}}{\text{ + }}$$     $${\text{2cos5}}{{\text{9}}^ \circ }$$  $$\frac{1}{{{\text{sin }}{{\left( {{{90}^ \circ } - {{59}^ \circ }} \right)}^ \circ }}}$$
$$\eqalign{ & \Rightarrow 3 + 2 \cr & \Rightarrow 5 \cr} $$

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

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

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

$$\eqalign{ & {\sin ^2}{5^ \circ } + {\sin ^2}{6^ \circ } + \,.....\,{\sin ^2}{84^ \circ } + {\sin ^2}{85^ \circ } \cr & {\text{Number of terms}} \cr & = \frac{{85 - 5}}{1} + 1 \cr & = 80 + 1 \cr & = 81 \cr} $$
  $$ = \left( {{{\sin }^2}{5^ \circ } + {{\sin }^2}{{85}^ \circ }} \right)\, + $$     $$\,\left( {{{\sin }^2}{6^ \circ } + {{\sin }^2}{{84}^ \circ }} \right)\, + $$     . . . . . 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 sin²θ or cos²θ,
then sum of series will always be half of number of terms.
$$ = \frac{{81}}{2} = 40\frac{1}{2}$$

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

Given, (sin²5° + sin²10° + sin²15° + . . . . . . + sin²85° ) + sin²90°
We know that sinθ = cos(90° - θ )
Therefore sin²85° = cos²(90° - 85°) = cos²5°
Similary sin²60° = cos²(90° - 60°) = cos²40°
And we also know that sin²θ + cos²θ = 1
There are 8 pair in given equation
sin²5° + cos²5° = 1
sin²10° + cos²10° = 1
. . . . . . . . . . . .
. . . . . . . . . . . .
sin²40° + cos²40° = 1
i.e. 8 + sin²45° + sin²90°
⇒ 8 + $$\frac{1}{2}$$ + 1 = $$9\frac{1}{2}$$

Short Trick:
 $$\left( {si{n^2}{5^ \circ }\, + \,{{\sin }^2}{{10}^ \circ }\, + \,{{\sin }^2}{{15}^ \circ }\, + \,.....\,{{\sin }^2}{{85}^ \circ }} \right)\, + $$         $$\,{\sin ^2}90$$
$$\eqalign{ & {\text{Number of terms}} \cr & = \left\{ {\left( {\frac{{85 - 5}}{5}} \right) + 1} \right\}{\text{ + 1}} \cr & {\text{ = }}\frac{{17}}{2}{\text{ + 1}} \cr & {\text{Sum of series}} = 9\frac{1}{2} \cr} $$

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

$$\eqalign{ & {\text{ta}}{{\text{n}}^2}\theta = 1 - {e^2} \cr & \therefore sec\theta + {\text{ta}}{{\text{n}}^3}\theta . {\text{cosec}}\theta \cr & \Rightarrow sec\theta + {\text{ta}}{{\text{n}}^2}\theta .{\text{tan}}\theta . {\text{cosec}}\theta \cr & \Rightarrow sec\theta + {\text{ta}}{{\text{n}}^2}\theta .\frac{{\sin \theta }}{{{\text{cos}}\theta }}.\frac{1}{{\sin \theta }} \cr & \Rightarrow sec\theta + {\text{ta}}{{\text{n}}^2}\theta .sec\theta \cr & \Rightarrow sec\theta \left( {1 + {\text{ta}}{{\text{n}}^2}\theta } \right) \cr & \Rightarrow \sqrt {1 + {\text{ta}}{{\text{n}}^2}\theta } .\left( {1 + {\text{ta}}{{\text{n}}^2}\theta } \right) \cr & \Rightarrow {\left( {1 + {\text{ta}}{{\text{n}}^2}\theta } \right)^{\frac{3}{2}}} \cr & \Rightarrow {\left( {1 + 1 - {e^2}} \right)^{\frac{3}{2}}} \cr & \Rightarrow {\left( {2 - {e^2}} \right)^{\frac{3}{2}}} \cr} $$

$$\eqalign{ & \frac{{{\text{cos }}\alpha }}{{{\text{cos }}\beta }} = a{\text{ }} \cr & \Rightarrow \cos {\text{ }}\alpha = a{\text{ }}\cos {\text{ }}\beta \cr & {\text{On squaring both sides}} \cr & {\cos ^2}\alpha = {a^2}{\cos ^2}\beta \cr & \Rightarrow 1 - {\sin ^2}\alpha = {a^2}\left( {1 - {{\sin }^2}\beta } \right)....(i) \cr & {\text{Again, }}\sin \alpha = {\text{ }}b\sin \beta \cr & {\text{Squaring both sides}} \cr & \Rightarrow {\sin ^2}\alpha = {\text{ }}{b^2}{\sin ^2}\beta \cr & {\text{Put the value of }}{\sin ^2}\alpha {\text{ in equation (i)}} \cr & \Rightarrow {\text{1}} - {b^2}{\sin ^2}\beta = {a^2} - {a^2}si{n^2}\beta \cr & \Rightarrow {a^2} - 1 = {a^2}si{n^2}\beta - {b^2}si{n^2}\beta \cr & \Rightarrow {a^2} - 1 = si{n^2}\beta \left( {{a^2} - {b^2}} \right) \cr & \Rightarrow si{n^2}\beta = \frac{{{a^2} - 1}}{{{a^2} - {b^2}}} \cr} $$

$$\eqalign{ & xsin\theta + y\cos \theta = 4 \cr & \underline {x\cos \theta - y\sin \theta = 2} {\text{ }}\,\,\,\,\left[ {{\text{Formula}}} \right] \cr & \left( {{x^2} + {y^2}} \right)\left( {{\text{co}}{{\text{s}}^2}\theta + {{\sin }^2}\theta } \right) = {4^2} + {2^2} \cr & \Rightarrow {x^2} + {y^2} = {a^2} + {b^2} \cr & \Rightarrow \left( {{x^2} + {y^2}} \right)\left( 1 \right) = 16 + 4 \cr & \Rightarrow {x^2} + {y^2} = 20 \cr} $$

  $$\left[ {\frac{{{\text{co}}{{\text{s}}^2}{\text{A}}{\text{.si}}{{\text{n}}^2}{\text{A}}\left( {{\text{sin A}} + {\text{cos A}}} \right)}}{{\left( {{\text{sin A}} - {\text{cos A}}} \right)}} + \frac{{{\text{si}}{{\text{n}}^2}{\text{A}}{\text{.co}}{{\text{s}}^2}{\text{A}}\left( {{\text{sin A}} - {\text{cos A}}} \right)}}{{\left( {{\text{sin A}} + {\text{cos A}}} \right)}}} \right]$$           $$\left[ {\frac{1}{{{\text{co}}{{\text{s}}^2}{\text{A}}}} - \frac{1}{{{\text{si}}{{\text{n}}^2}{\text{A}}}}} \right]$$
  $$ \Rightarrow \left[ {\frac{{{{\left( {{\text{sin A}} + {\text{cos A}}} \right)}^2} + {{\left( {{\text{sin A}} - {\text{cos A}}} \right)}^2}}}{{\left( {{\text{sin A}} - {\text{cos A}}} \right)\left( {{\text{sin A}} + {\text{cos A}}} \right)}}} \right]$$       $$\left( {{\text{si}}{{\text{n}}^2}{\text{ A}} - {\text{co}}{{\text{s}}^2}{\text{ A}}} \right)$$
$$\eqalign{ & \Rightarrow 2\left( {{\text{si}}{{\text{n}}^2}{\text{ A}} + {\text{co}}{{\text{s}}^2}{\text{ A}}} \right) \cr & \Rightarrow 2 \cr} $$

According to the question,
  $$2{\operatorname{cosec} ^2}{23^ \circ }{\text{ co}}{{\text{t}}^2}{67^ \circ } - {\sin ^2}{23^ \circ } - $$       $${\sin ^2}{67^ \circ } - $$     $${\text{co}}{{\text{t}}^2}{67^ \circ }$$
  $$ \Rightarrow 2{\operatorname{cosec} ^2}{23^ \circ }{\text{ co}}{{\text{t}}^2}\left( {{{90}^ \circ } - {{23}^ \circ }} \right) - $$       $${\sin ^2}{23^ \circ } - $$   $${\sin ^2}\left( {{{90}^ \circ } - {{23}^ \circ }} \right) - $$     $${\text{co}}{{\text{t}}^2}{67^ \circ }$$
  $$ \Rightarrow 2{\operatorname{cosec} ^2}{23^ \circ }{\text{ ta}}{{\text{n}}^2}{23^ \circ } - $$     $$\left( {{{\sin }^2}{{23}^ \circ } + co{s^2}{{23}^ \circ }} \right) - $$   $${\text{co}}{{\text{t}}^2}{67^ \circ }$$
$$\eqalign{ & \Rightarrow \frac{2}{{{\text{co}}{{\text{s}}^2}{{23}^ \circ }}} - 1 - {\text{co}}{{\text{t}}^2}{67^ \circ } \cr & \Rightarrow 2se{c^2}{23^ \circ } - {\text{1}} - {\text{co}}{{\text{t}}^2}\left( {{{90}^ \circ } - {{23}^ \circ }} \right) \cr & \Rightarrow 2se{c^2}{23^ \circ } - {\text{1}} - {\text{ta}}{{\text{n}}^2}{23^ \circ } \cr & \Rightarrow 2se{c^2}{23^ \circ } - \left( {{\text{1}} + {\text{ta}}{{\text{n}}^2}{{23}^ \circ }} \right) \cr & \Rightarrow 2se{c^2}{23^ \circ } - {\sec ^2}{23^ \circ } \cr & \Rightarrow se{c^2}{23^ \circ } \cr} $$

$$\eqalign{ & \frac{{{\text{tan}}\theta }}{{1 - {\text{cot}}\theta }} + \frac{{{\text{cot}}\theta }}{{1 - {\text{tan}}\theta }} \cr & \Rightarrow \frac{{{\text{tan}}\theta }}{{1 - \frac{1}{{\tan \theta }}}} + \frac{{\frac{1}{{\tan \theta }}}}{{1 - {\text{tan}}\theta }} \cr & \Rightarrow \frac{{{\text{ta}}{{\text{n}}^2}\theta }}{{{\text{tan}}\theta - 1}} + \frac{1}{{{\text{tan}}\theta \left( {1 - \tan \theta } \right)}} \cr & \Rightarrow \frac{{{\text{tan}}^2\theta }}{{{\text{tan}}\theta - 1}} - \frac{1}{{{\text{tan}}\theta \left( {\tan \theta - 1} \right)}} \cr & \Rightarrow \frac{{{\text{ta}}{{\text{n}}^3}\theta - 1}}{{{\text{tan}}\theta \left( {\tan \theta - 1} \right)}} \cr & \Rightarrow \frac{{\left( {{\text{tan}}\theta - 1} \right)\left( {{\text{ta}}{{\text{n}}^2}\theta + {\text{tan}}\theta + {\text{1}}} \right)}}{{{\text{tan}}\theta \left( {\tan \theta - 1} \right)}} \cr & \Rightarrow \frac{{{\text{ta}}{{\text{n}}^2}\theta + {\text{tan}}\theta + {\text{1}}}}{{{\text{tan}}\theta }} \cr & \Rightarrow {\text{tan}}\theta + \cot \theta + {\text{1}} \cr} $$

$$\eqalign{ & \frac{{\cot {{30}^ \circ } - \cot {{75}^ \circ }}}{{\tan {{15}^ \circ } - \tan {{60}^ \circ }}} \cr & = \frac{{\tan {{60}^ \circ } - \tan {{15}^ \circ }}}{{\tan {{15}^ \circ } - \tan {{60}^ \circ }}} \cr & = \frac{{ - \left( {\tan {{15}^ \circ } - \tan {{60}^ \circ }} \right)}}{{\tan {{15}^ \circ } - \tan {{60}^ \circ }}} \cr & = - 1 \cr} $$

$$\eqalign{ & \left( {{\text{se}}{{\text{c}}^4}\theta - {\text{ta}}{{\text{n}}^4}\theta } \right) \cr & \Rightarrow \left( {{\text{se}}{{\text{c}}^2}\theta - {\text{ta}}{{\text{n}}^2}\theta } \right)\left( {{\text{se}}{{\text{c}}^2}\theta + {\text{ta}}{{\text{n}}^2}\theta } \right) \cr & \Rightarrow 1 \times \left( {{\text{se}}{{\text{c}}^2}\theta + {\text{ta}}{{\text{n}}^2}\theta } \right)[1 + {\text{ta}}{{\text{n}}^2}\theta = {\text{se}}{{\text{c}}^2}\theta ] \cr & \Rightarrow 1 \times \frac{7}{{12}} \cr & \Rightarrow \frac{7}{{12}} \cr} $$

$$\eqalign{ & \sin \theta = 0.7 \cr & \Rightarrow {\sin ^2}\theta + {\text{co}}{{\text{s}}^2}\theta = 1 \cr & \Rightarrow {\left( {0.7} \right)^2} + {\text{co}}{{\text{s}}^2}\theta = 1 \cr & \Rightarrow 0.49 + {\text{co}}{{\text{s}}^2}\theta = 1 \cr & \Rightarrow {\text{co}}{{\text{s}}^2}\theta = 1 - 0.49 \cr & \Rightarrow {\text{cos}}\theta = \sqrt {0.51} \cr} $$

$$\eqalign{ & {\text{2}}\sin \theta + {\text{cos}}\theta = \frac{7}{3} \cr & \Rightarrow \left( {{\text{ta}}{{\text{n}}^2}\theta - {\text{se}}{{\text{c}}^2}\theta } \right) \cr & \Rightarrow \left( {{{\sec }^2}\theta - 1 - {\text{se}}{{\text{c}}^2}\theta } \right) \cr & \,\,\,\,\,\,\,\,\,\,\,\left[ {\because 1 + {\text{ta}}{{\text{n}}^2}\theta = {{\sec }^2}\theta } \right] \cr & \Rightarrow - 1 \cr} $$

$$\eqalign{ & {\sec ^2}\theta - \frac{{{{\sin }^2}\theta - 2{{\sin }^4}\theta }}{{{\text{2co}}{{\text{s}}^4}\theta - {\text{co}}{{\text{s}}^2}\theta }} \cr & \Rightarrow {\sec ^2}\theta - \frac{{{{\sin }^2}\theta \left( {1 - 2{{\sin }^2}\theta } \right)}}{{{\text{co}}{{\text{s}}^2}\theta \left( {{\text{2co}}{{\text{s}}^2}\theta - 1} \right)}} \cr & \left[ {{\text{co}}{{\text{s}}^2}\theta - {{\sin }^2}\theta = 2{\text{co}}{{\text{s}}^2}\theta - 1 = 1 - 2{{\sin }^2}\theta } \right] \cr & \Rightarrow {\sec ^2}\theta - {\text{ta}}{{\text{n}}^2}\theta \cr & \Rightarrow 1 \cr} $$

$$\eqalign{ & {\text{co}}{{\text{s}}^2}\theta + {\text{co}}{{\text{s}}^4}\theta = 1 \cr & \Rightarrow {\text{co}}{{\text{s}}^4}\theta = 1 - {\cos ^2}\theta \cr & \Rightarrow {\text{co}}{{\text{s}}^4}\theta = {\sin ^2}\theta \cr & \Rightarrow {\cos ^2}\theta .{\cos ^2}\theta = {\sin ^2}\theta \cr & \Rightarrow {\cos ^2}\theta = \frac{{{{\sin }^2}\theta }}{{{{\cos }^2}\theta }} \cr & \Rightarrow {\text{co}}{{\text{s}}^2}\theta = {\text{ta}}{{\text{n}}^2}\theta \cr & \Rightarrow {\text{ta}}{{\text{n}}^2}\theta + {\text{ta}}{{\text{n}}^4}\theta \cr & \Rightarrow {\text{co}}{{\text{s}}^2}\theta + {\text{co}}{{\text{s}}^4}\theta = 1 \cr} $$

$$\eqalign{ & {\text{co}}{{\text{s}}^4}\theta - {\sin ^4}\theta = \frac{2}{3} \cr & \Rightarrow \left( {{\text{co}}{{\text{s}}^2}\theta - {{\sin }^2}\theta } \right)\left( {{\text{co}}{{\text{s}}^2}\theta + {{\sin }^2}\theta } \right) = \frac{2}{3} \cr & \Rightarrow {\text{co}}{{\text{s}}^2}\theta - {\sin ^2}\theta = \frac{2}{3} \cr & \Rightarrow 1 - {\sin ^2}\theta - {\sin ^2}\theta = \frac{2}{3} \cr & \Rightarrow 1 - 2{\sin ^2}\theta = \frac{2}{3} \cr} $$

$$\eqalign{ & \frac{{{\text{sin A}}}}{{1 + \cos {\text{ A}}}}{\text{ + }}\frac{{{\text{sin A}}}}{{1 - \cos {\text{ A}}}} \cr & \Rightarrow \frac{{{\text{sin A}}\left( {1 - \cos {\text{ A}}} \right) + {\text{sin A}}\left( {1 + \cos {\text{ A}}} \right)}}{{\left( {1 + \cos {\text{ A}}} \right)\left( {1 - \cos {\text{ A}}} \right)}} \cr & \Rightarrow \frac{{{\text{sin A}} - {\text{sin A}}{\text{.cosA}} + {\text{sin A}} + {\text{sin A}}{\text{.cosA}}}}{{{\text{1}} - {\text{co}}{{\text{s}}^2}{\text{A}}}} \cr & \Rightarrow \frac{{2{\text{sin A}}}}{{{{\sin }^2}{\text{A}}}} \cr & \Rightarrow 2{\text{ cosec A}} \cr} $$

$$\eqalign{ & \cos \pi x = {x^2} - x + \frac{5}{4} \cr & = {x^2} - 2 \times x \times \frac{1}{2} + \frac{1}{4} - \frac{1}{4} + \frac{5}{4} \cr & = {\left( {x - \frac{1}{2}} \right)^2} + 1 > 1 \cr & = - 1 \leqslant \cos x \leqslant 1 \cr} $$
∴ So, value of x is none of the above

$$\eqalign{ & {\bf{Shortcut \,\, method:}} \cr & = \left( {1 + \cot \theta - \operatorname{cosec} \theta } \right)\left( {1 + \tan \theta + \sec \theta } \right) \cr & \left[ {put,\theta = {{45}^ \circ }} \right] \cr} $$
  $$ = \left( {1 + \cot {{45}^ \circ } - \operatorname{cosec} {{45}^ \circ }} \right)$$     $$\left( {1 + \tan {{45}^ \circ } + \sec {{45}^ \circ }} \right)$$
$$\eqalign{ & = \left( {1 + 1 - \sqrt 2 } \right)\left( {1 + 1 + \sqrt 2 } \right) \cr & = \left( {2 - \sqrt 2 } \right)\left( {2 + \sqrt 2 } \right) \cr & = \left[ {{2^2} - {{\left( {\sqrt 2 } \right)}^2}} \right]\left[ {\left( {a - b} \right)\left( {a + b} \right) = {a^2} - {b^2}} \right] \cr & = 4 - 2 \cr & = 2 \cr} $$

$$\eqalign{ & x = a{\text{ }}\sec \theta .\cos \phi \cr & y = b{\text{ }}\sec \theta .sin\phi \cr & z = c{\text{ tan}}\theta \cr & \frac{x}{a} = \sec \theta .\cos \phi \cr & \frac{y}{b} = \sec \theta .sin\phi \cr & \frac{z}{c} = {\text{tan}}\theta \cr & \therefore \frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} - \frac{{{z^2}}}{{{c^2}}} \cr & \Rightarrow {\sec ^2}\theta .{\cos ^2}\phi + {\sec ^2}\theta .si{n^2}\phi - {\text{ta}}{{\text{n}}^2}\theta \cr & \Rightarrow {\sec ^2}\theta \left( {{{\cos }^2}\phi + si{n^2}\phi } \right) - {\text{ta}}{{\text{n}}^2}\theta \cr & \Rightarrow {\sec ^2}\theta - {\text{ta}}{{\text{n}}^2}\theta \cr & \Rightarrow 1 \cr} $$

$$\eqalign{ & \frac{5}{{{\text{se}}{{\text{c}}^2}\theta }}{\text{ + }}\frac{2}{{1 + {\text{co}}{{\text{t}}^2}\theta }}{\text{ + 3}}{\sin ^2}\theta \cr & = 5{\cos ^2}\theta + \frac{2}{{{\text{cose}}{{\text{c}}^2}\theta }} + 3{\sin ^2}\theta \cr & = 5{\text{co}}{{\text{s}}^2}\theta + 2{\sin ^2}\theta + 3{\sin ^2}\theta \cr & = 5\left( {{\text{co}}{{\text{s}}^2}\theta + {{\sin }^2}\theta } \right) \cr & \left( {\because {{\sin }^2}\theta + {\text{co}}{{\text{s}}^2}\theta = 1} \right) \cr & = 5 \times 1 \cr & = 5 \cr} $$

$$\eqalign{ & \left( {\frac{1}{{\cos \theta }} + \frac{1}{{\cot \theta }}} \right){\text{ }}\left( {\frac{1}{{\cos \theta }} - \frac{1}{{\cot \theta }}} \right) \cr & = \left( {\sec \theta + \tan \theta } \right)\left( {\sec \theta - \tan \theta } \right) \cr & = {\sec ^2}\theta - {\tan ^2}\theta \left[ {1 + {{\tan }^2}\theta = {{\sec }^2}\theta } \right] \cr & = 1 \cr} $$

$$\frac{{3\sin \theta + 2{\text{cos}}\theta }}{{3\sin \theta - 2{\text{cos}}\theta }}$$
Divide numerator & denominator by cosθ
$$\eqalign{ & = \frac{{\frac{{3\sin \theta }}{{\cos \theta }} + \frac{{2\cos \theta }}{{\cos \theta }}}}{{\frac{{3\sin \theta }}{{\cos \theta }} - \frac{{2\cos \theta }}{{\cos \theta }}}}\left[ {\frac{{\sin \theta }}{{\cos \theta }} = \tan \theta } \right] \cr & = \frac{{3\tan \theta + 2}}{{3\tan \theta - 2}} \cr & {\text{Put value of tan}}\theta \cr & = \frac{{3 \times \frac{4}{3} + 2}}{{3 \times \frac{4}{3} - 2}} \cr & = \frac{6}{2} \cr & = 3 \cr} $$

(secA - cosA)² + (cosecA - sinA)² - (cotA - tanA)²
= (sec²A + cos²A - 2secA.cosA) + (coses²A + sin²A - 2cosecA.sinA) - (cot²A + tan²A - 2cotA.tanA)
= sec²A - tan²A + cos²A + sin²A + coses²A - cot²A - 2
= 3 - 2
= 1

Alternate shortcut method:
(secA - cosA)² + (cosecA - sinA)² - (cotA - tanA)²
Put θ = 45°
= (sec45° - cos45°)² + (cosec45° - sin45°)² - (cot45° - tan45°)²
$$\eqalign{ & = {\left( {\sqrt 2 - \frac{1}{{\sqrt 2 }}} \right)^2} + {\left( {\sqrt 2 - \frac{1}{{\sqrt 2 }}} \right)^2} - {\left( {1 - 1} \right)^2} \cr & = \frac{1}{2} + \frac{1}{2} - 0 \cr & = 1 \cr} $$