Solution:
$$\eqalign{ \frac{{{{\left[ {\tan \left( {{{90}^ \circ } - A} \right) + \cot \left( {{{90}^ \circ } - A} \right)} \right]}^2}}}{{\left[ {2{{\sec }^2}\left( {{{90}^ \circ } - 2A} \right)} \right]}} \cr {\text{By putting }}A = {45^ \circ } \cr \Rightarrow \frac{{{{\left[ {\tan \left( {{{90}^ \circ } - {{45}^ \circ }} \right) + \cot \left( {{{90}^ \circ } - {{45}^ \circ }} \right)} \right]}^2}}}{{\left[ {2{{\sec }^2}\left( {{{90}^ \circ } - 2 \times {{45}^ \circ }} \right)} \right]}} \cr \Rightarrow \frac{{{{\left[ {\tan {{45}^ \circ } + \cot {{45}^ \circ }} \right]}^2}}}{{2{{\sec }^2}\left( {{{90}^ \circ } - {{90}^ \circ }} \right)}} \cr \Rightarrow \frac{{{{\left[ {1 + 1} \right]}^2}}}{{2{{\sec }^2}{0^ \circ }}} \cr \Rightarrow \frac{4}{{2 \times 1}} \cr \Rightarrow 2 \cr} $$

Solution:
(x + 5)° + (2x - 3)° + (3x + 4)° = 180°
(Sum of all angles in triangle is 180°)
⇒ 6x + 6° = 180°
⇒ (x + 1) = 30°
⇒ x = 29°

Solution:
$$\eqalign{ {\text{7}}{\sin ^2}\theta + 3{\cos ^2}\theta = 4 \cr \Rightarrow {\text{7}}{\sin ^2}\theta + 3\left( {1 - {{\sin }^2}\theta } \right) = 4 \cr \Rightarrow {\text{7}}{\sin ^2}\theta + 3 - 3{\sin ^2}\theta = 4 \cr \Rightarrow 4{\sin ^2}\theta = 1 \cr \Rightarrow {\sin ^2}\theta = \frac{1}{4} \cr \Rightarrow \sin \theta = \frac{1}{2} \cr \Rightarrow \sin \theta = \sin {30^ \circ } \cr \Rightarrow \theta = {30^ \circ } \cr \tan {30^ \circ } = \frac{1}{{\sqrt 3 }} \cr \cr {\bf{Alternate:}} \cr {\text{Put}}\theta = {30^ \circ } \cr {\text{7}} \times \sin^2 {30^ \circ } + 3{\cos ^2}{30^ \circ } = 4 \cr \Rightarrow 7 \times \frac{1}{4} + 3 \times \frac{3}{4} = 4 \cr \Rightarrow \frac{7}{4} + \frac{9}{4} = 4 \cr \Rightarrow \frac{{16}}{4} = 4 \cr \Rightarrow 4 = 4\left( {{\text{Satisfied}}} \right) \cr \therefore \tan {30^ \circ } = \frac{1}{{\sqrt 3 }} \cr} $$

Solution:
$$\eqalign{ = \frac{{2\sin \theta }}{{\cos \theta \left( {1 + {\text{ta}}{{\text{n}}^2}\theta } \right)}} \cr = \frac{{2{\text{tan}}\theta }}{{1 + {\text{ta}}{{\text{n}}^2}\theta }} \cr = \sin2\theta \cr} $$

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

Solution:
$$\eqalign{ \frac{{\sec \theta \left( {1 - \sin \theta } \right)\left( {\sin \theta + \cos \theta } \right)\left( {\sec \theta + \tan \theta } \right)}}{{\sin \theta \left( {1 + \tan \theta } \right) + \cos \theta \left( {1 + \cot \theta } \right)}} \cr = \frac{{\left( {\sec \theta - \tan \theta } \right)\left( {\sin \theta + \cos \theta } \right)\left( {\sec \theta + \tan \theta } \right)}}{{\sin \theta \left( {1 + \tan \theta } \right) + \cos \theta \left( {\frac{{1 + \tan \theta }}{{\tan \theta }}} \right)}} \cr = \frac{{\left( {{{\sec }^2}\theta - {{\tan }^2}\theta } \right)\left( {\sin \theta + \cos \theta } \right)}}{{\left( {1 + \tan \theta } \right)\left( {\sin \theta + \frac{{{{\cos }^2}\theta }}{{\sin \theta }}} \right)}} \cr = \frac{{1\left( {\sin \theta + \cos \theta } \right)}}{{\left( {1 + \tan \theta } \right)\left( {\frac{1}{{\sin \theta }}} \right)}} \cr = \frac{{\left( {\sin \theta + \cos \theta } \right)}}{{\left( {\cos \theta + \sin \theta } \right)}}\sin \theta .\cos \theta \cr = \sin \theta .\cos \theta \cr} $$

Solution:
$$\cos \theta = \frac{{15 \to {\text{Base}}}}{{17 \to {\text{Hypo}}}}$$

$$\eqalign{ {\text{Perpendicular = 8}} \cr \Rightarrow {\text{cot}}\left( {{{90}^ \circ } - \theta } \right) \cr \Rightarrow {\text{tan}}\theta = \frac{8}{{15}}\left[ {\therefore \tan \theta = \frac{{\text{P}}}{{\text{B}}}} \right] \cr} $$