We have
L.H.S. =\({cos9x - cos 5x\over sin17x - sin 3x}\)
\(={-2sin({9x+5x\over 2})sin ({9x+5x\over 2})\over 2cos({17x+3x\over2})sin({17x+3x\over2})}\)
\(\left[\begin{array}{c} \because \cos \mathrm{C}-\cos \mathrm{D}=-2 \sin \frac{\mathrm{C}+\mathrm{D}}{2} \sin \frac{\mathrm{C}-\mathbf{D}}{2} \\ \sin \mathrm{C}-\sin \mathrm{D}=2 \cos \frac{\mathrm{C}+\mathrm{D}}{2} \sin \frac{\mathrm{C}-\mathrm{D}}{2} \end{array}\right]\)
\(={-2sin7xsin2x\over 2cos10xsin7x}=-{sin 2x\over cos 10x} = R.H.S\)

Here sec x=\(13\over 5\)
Now sec²x = 1+ tan² x
\(\Rightarrow ({13\over 5})^2=1+tan^2x\)
\(\Rightarrow tan^2 x={169\over25}-1\)
\(\Rightarrow tan^2x={144\over 25}\)
\(\Rightarrow tan x=\pm {12\over 5}\)
But x lies in fourth quadrant.
\(\therefore tan x={-12\over5}\)
\(cot \ x={1\over tan \ x}={-5\over 12}\)
\(cos x={1\over sec \ x}={5\over 13}\)
Also sin² x + cos² x = 1
\(\Rightarrow sin ^2x+({5\over 13})^2=1\)
\(\Rightarrow sin ^2x=1-{25\over 169}\)
\(\Rightarrow sin^2 x={144\over 169}\)
\(\Rightarrow sin \ x =\pm {12\over 13}\)
But x lies in fourth quadrant.
\(\therefore sin x={-12\over13}\)

\(LHS=(cosA+cosB)+cosC\)
\(=2cos\left( \frac { A+B }{ 2 } \right) cos\left( \frac { A-B }{ 2 } \right) +1-2sin^{ 2 }\frac { c }{ 2 }\)
\(=2cos\left( \frac { \pi }{ 2 } -\frac { c }{ 2 } \right) cos\left( \frac { A-B }{ 2 } \right) -2sin^{ 2 }\frac { C }{ 2 } +1\)
\(=2sin\frac { C }{ 2 } cos\left( \frac { A-B }{ 2 } \right) -2sin^{ 2 }\frac { C }{ 2 } +1\)
\(=2sin\frac { C }{ 2 } \left[ cos\left( \frac { A-B }{ 2 } \right) -sin\frac { C }{ 2 } \right] +1\)
\(=2sin\frac { C }{ 2 } \left[ cos\left( \frac { A-B }{ 2 } \right) -cos\left( \frac { A+B }{ 2 } \right) \right] +1\)
\(=2sin\frac { C }{ 2 } \left[ -2sin\frac { A }{ 2 } sin\left( \frac { -B }{ 2 } \right) \right] +1\)
\(=4sin\frac { C }{ 2 } sin\frac { A }{ 2 } sin\frac { B }{ 2 } +1\)

\(cos^{ 4 }\frac { \pi }{ 8 } +cos^{ 4 }\frac { 3\pi }{ 8 } +cos^{ 4 }\frac { 5\pi }{ 8 } +cos^{ 4 }\frac { 7\pi }{ 8 } \)
\(=cos^{ 4 }\frac { \pi }{ 8 } +cos^{ 4 }\frac { 3\pi }{ 8 } +cos^{ 4 }\left( \pi -\frac { 3\pi }{ 8 } \right) +cos^{ 4 }\left( \pi -\frac { \pi }{ 8 } \right) \)
\(= cos^{ 4 }\frac { \pi }{ 8 } +cos^{ 4 }\frac { 3\pi }{ 8 } +cos^{ 4 }\frac { \pi }{ 8 } \)
\(= 2\left[ cos^{ 4 }\frac { \pi }{ 8 } +cos^{ 4 }\frac { 3\pi }{ 8 } \right] =2\left[ cos^{ 4 }\frac { \pi }{ 8 } +cos^{ 4 }\left( \frac { \pi }{ 2 } -\frac { \pi }{ 8 } \right) \right] \)
\( =2-\left( sin\frac { 2\pi }{ 8 } \right) ^{ 2 }=\frac { 3 }{ 2 } \)

We know that 3x = 2x + x
Therefore, tan 3x = tan (2x + x)
or \(\tan 3 x=\frac{\tan 2 x+\tan x}{1-\tan 2 x \tan x}\)
or tan 3x – tan 3x tan 2x tan x = tan 2x + tan x
or tan 3x – tan 2x – tan x = tan 3x tan 2x tan x
or tan 3x tan 2x tan x = tan 3x – tan 2x – tan x.

We have
L.H.S. = sin 3x + sin 2x - sin x.
= (sin 3x - sin x) + sin 2x
\(=[2cos({3x+x\over2}) sin({3x+x\over2})]+2sin x cos x\)
\([\because \ sin C- sin D=2cos {C+D\over2}.sin{C-D\over2}]\)
= 2 cos 2x sin x + 2 sin x cos x
= 2 sin x [cos 2x + cos x]
= 2 sin x\([2cos({2cos ({2x+x\over2})}cos({2x+x\over2})]\)
\(=2sin x[2cos({3x\over2})cos({x\over2})]\)
= 4 sin x cos \({x\over2}\) cos \({3x\over2}\) =R.H.S

We have
L.H.S.= sin x + sin 3x + sin 5x + sin 7x
= (sin 7x + sin x) + (sin 5x + sin 3x)
\(=[2sin({7x+x\over2})cos({7x-x\over2})]+[2sin({5x+3x\over2})cos({5x-3x\over2})]\)
= 2 sin 4x cos 3x + 2 sin 4x cos x
\([\because \ sinC+sin D=2sin{C+D\over2}.cos{C-D\over2}]\)
= 2 sin 4x [cos 3x + cos x]
= 2 sin 4x\([2cos({3x+x\over2})cos({3x-x\over2})]\)
\([\because \ cos C+cosD=2cos {{C+D\over2}.cos{C-D\over2}}]\)
= 2 sin 4x [2 cos 2x cos x]
= 4 cos x cos 2x sin 4x = R.H.S.

We have 
L.H.S.\(={sin \ x - sin \ y\over cos \ x+ cos \ y }\)
\(={2cos({x+y\over 2})sin({x-y \over 2})\over {2cos({x+y\over 2}) cos({x-y \over 2})}}\)
\(\left[\begin{array}{l} \because \sin \mathrm{C}+\sin \mathrm{D}=2 \sin \left(\frac{\mathrm{C}+\mathbf{D}}{2}\right) \cos \left(\frac{\mathbf{C}-\mathbf{D}}{2}\right) \\ \cos \mathrm{C}+\cos \mathrm{D}=2 \cos \left(\frac{\mathbf{C}+\mathrm{D}}{2}\right) \cos \left(\frac{\mathrm{C}-\mathrm{D}}{2}\right) \end{array}\right]\)
\(={sin {x-y\over 2}\over cos {x-y\over2}}=tan {x-y\over2}\)
=R.H.S

(i) sin 75° = sin (45° + 30°)
= sin 45° cos 30° + cos 45° sin 30°
[\(\because\) sin (A+ B) = sin A cos B + sin B cos A]
\(={1\over\sqrt{2}}\times{\sqrt{3}\over2}+{1\over\sqrt{2}}\times{1\over2}\)
\(={\sqrt{3}\over2\sqrt{2}}+{1\over2\sqrt{2}}={\sqrt{3}+1\over2\sqrt{2}}\)
(ii) \({ tan15 }^{ 0 }=tan({ 60 }^{ 0 }-{ 45 }^{ 0 })\)
\(=\frac { { tan60 }^{ 0 }-{ tan45 }^{ 0 } }{ 1+{ tan60 }^{ 0 }{ tan45 }^{ 0 } } \left[ tan(A-B)=\frac { tanA-tanB }{ 1+tanAtanB } \right]\)
\(=\frac { \sqrt { 3 } -1 }{ 1+\sqrt { 3 } .1 } \left[ { tan45 }^{ 0 }=1\quad and\quad { tan60 }^{ 0 }=\sqrt { 3 } \right]\)
\(=\frac { \sqrt { 3 } -1 }{ 1+\sqrt { 3 } } =\frac { \sqrt { 3 } -1 }{ \sqrt { 3 } +1 } =\frac { \sqrt { 3 } -1 }{ \sqrt { 3 } +1 } \times \frac { \sqrt { 3 } -1 }{ \sqrt { 3 } -1 } \left[ on\quad rationalisation \right] \)
\(=\frac { { \left( \sqrt { 3 } -1 \right) }^{ 2 } }{ { \sqrt { 3 } }^{ 2 }-\left( { 1 }^{ 2 } \right) } \left[ (a-b)(a+b)={ a }^{ 2 }-{ b }^{ 2 } \right] \)
\(=\frac { { \sqrt { 3 } }^{ 2 }+{ 1 }^{ 2 }-2\sqrt { 3 } }{ 3-1 } \left[ { \left( a-b \right) }^{ 2 }={ a }^{ 2 }{ +b }^{ 2 }-2ab \right] \)
\(=\frac { { \sqrt { 3 } }^{ 2 }+{ 1 }^{ 2 }-2\sqrt { 3 } }{ 3-1 } =\frac { 4-2\sqrt { 3 } }{ 2 } =\frac { 2(2-\sqrt { 3 } ) }{ 2 } =(2-\sqrt { 3 } )\)

To convert radian into degree measures,
we multiply by \(180\over \pi\)
We know that π radian = 180°
\(\therefore ({11\over 16})^C=12^o({2\over 5}\times 60)^o=({{11\over 16}\times{180\times7 \over 22 }})^o\)
\(=({315\over 8})^o=({39{5\over 8}})^o\)
\(=39^o({3\over 8}\times60)'=39^o({45\over 2})'\)
\(=39^o(22{1\over 2})'=39^o22'({1\over 2}\times 60)'\)
= 39^o 22' 30"
(ii) (-4)
We know that π radian = 180°
=\(-(4\times {180\over \pi})^o\)
\(=-(4\times {180\times 7\over22})^C=-({2520\over 11})^o\)
\(=-({229{1\over11}})^o=-229^o({1\over 11}\times 60)'\)
\(=-229^o(5{5\over 11})'=-229^o5'({5\over11}\times 60)''\)
\(=-229^o5'27{3\over 11}''\)
(iii) We know that π radian = 180°
\(({5\pi\over 3})^C=({5\pi\over3}\times {180\over \pi})^o=300^o\)
(iv) We know that π radian = 180°
\(({7\pi\over 6})^C=({7\pi\over 6}\times {180\over \pi})^o=210^o\)

We know that values of cos x repeats after an interval of 2\(\pi\) or 360°.
Therefore, cos (–1710°) = cos (–1710° + 5 × 360°)
= cos (–1710° + 1800°) = cos 90° = 0.

We have
L.H.S.= (cos x + cosy)² + (sin x - sin y)²\({x+y\over2}\)
\(=[2cos({x+y\over2})cos ({{x+y\over2}})]^2\)+\([2cos({x+y\over2})sin({x+y\over2})]^2\)
\(=4cos^2({x+y\over2})cos^2({x-y\over2})+4cos^2({x+y\over2})sin^2({x-y\over2})\)
\(=4cos^2({x+y\over2})[cos^2({x-y\over2})+sin^2({x-y\over2})]\)
\(=4cos^2({x+y\over2})=R.H.S\)

We have
L.H.S. = sin 2x + 2 sin 4x + sin 6x
= [sin 4x+ sin 2x] + [sin 6x + sin 4x]
= 2 sin\(({4x+2x\over 2})cos({4x-2x\over2})\)+\(2 sin ({6x+4x\over 2})cos({6x-4x\over2})\)
\([\because sin C+sin D=2sin({C+D\over 2})cos({C-D\over 2})]\)
= 2 sin 2x cos x + 2 sin 5x cos x
= 2 cos x [sin 3x + sin 5x]
\(=2cos \ x[2sin({5x+3x\over2}).cos({5x-3x\over2})]\)
= 2 cos x [2 sin 4x . cos x]
= 4 cos²x sin 4x = R.H.S.

tan \({19\pi\over 3}\)= tan \({19\over 3}\)x180
= tan 1140°
tan 1140° = tan (3 x 360° + 60°)
= tan 60° =\(\sqrt{3}\)

Hint Given, AB = 20 cm, AC= 12 cm

\(\therefore B C  =\sqrt{20^2-12^2}=16 \mathrm{cm} \)
\(\tan \gamma  =\frac{A C}{B C}=\frac{12}{16}=\frac{3}{4}\)
\([\triangle D E A \cong \triangle D E B\) by SAS property]
\(\gamma =\beta \)
\(\cos \gamma =\frac{B D}{B E} \Rightarrow B E=\frac{10}{4 / 5}=\frac{25}{2} \)
\(\tan \gamma =\frac{D E}{B D} \Rightarrow D E=\frac{10}{4 / 3}=\frac{15}{2} \)
\(C E =B C-B E=16-\frac{25}{2}=\frac{7}{2} \)
\(\tan \alpha =\frac{C E}{A C}=\frac{7}{24}, \cos \alpha=\frac{24}{25}, \sin \alpha=\frac{7}{25} \)
\(\tan \theta =\tan \left(\frac{\pi}{2}-\beta\right)=\cot \beta=\cot \gamma=\frac{4}{3}, \)
\(\cos \theta =\frac{3}{5}, \sin \theta=\frac{4}{5}\)
(i) (d) Use formula cos (A+B)= cos A cos B-sin A sin B
(ii) (c) Use formula \(\tan 2 A=\frac{2 \tan A}{1-\tan ^2 A}\)
(iii) (c) Since, \( A E>C E \ \delta>\alpha \)
and \(A D>D E  \beta<\theta \)
\( \gamma=\beta \quad \gamma<\theta\)
\( \therefore \gamma+\delta<\alpha+\theta\)
(iv) (d) Use formula
-2 sin A sin B=cos (A+B)-cos (A-B)
(v) (c)