We know that
sin (x + y) = sin x cos y + cos x sin y ... (1)
Now \(\cos ^{2} x=1-\sin ^{2} x=1-\frac{9}{25}=\frac{16}{25}\)
Therefore \(\cos x=\pm \frac{4}{5}\)
Since x lies in second quadrant, cos x is negative.
Hence \(\cos x=- \frac{4}{5}\)
Now sin²y = 1 – cos²y = 1 \(-\frac{144}{169}=\frac{25}{169}\)
i.e. \(\sin y=\pm \frac{5}{13}\)
Since y lies in second quadrant, hence sin y is positive. Therefore,\(\sin y= \frac{5}{13}\) Substituting the values of sin x, sin y, cos x and cos y in (1), we get
\(\sin (x+y)=\frac{3}{5} \times\left(-\frac{12}{13}\right)+\left(-\frac{4}{5}\right) \times \frac{5}{13}=-\frac{36}{65}-\frac{20}{65}=-\frac{56}{65}\)

We have
L.H.S. \(=\frac{\sin 5 x-2 \sin 3 x+\sin x}{\cos 5 x-\cos x}=\frac{\sin 5 x+\sin x-2 \sin 3 x}{\cos 5 x-\cos x} \)
\(=\frac{2 \sin 3 x \cos 2 x-2 \sin 3 x}{-2 \sin 3 x \sin 2 x}=-\frac{\sin 3 x(\cos 2 x-1)}{\sin 3 x \sin 2 x} \)
\(=\frac{1-\cos 2 x}{\sin 2 x}=\frac{2 \sin ^{2} x}{2 \sin x \cos x}=\tan x=\mathrm{R} . \mathrm{H.S}\)

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 : \(3 \sin \frac{\pi}{6} \sec \frac{\pi}{3}-4 \sin \frac{5 \pi}{6} \cot \frac{\pi}{4}\)
\(=3 \times \frac{1}{2} \times 2-4 \sin \left(\pi-\frac{\pi}{6}\right) \times 1=3-4 \sin \frac{\pi}{6} \)
\(=3-4 \times \frac{1}{2}=1=\mathrm{R} . \mathrm{H.S} .\)

We have
L.H.S. = 2cos\({\pi\over13}\)cos\({9\pi\over13}\)+cos\({3\pi\over13}\)+cos\({5\pi\over13}\)
\(=cos({9\pi\over13}+{\pi\over13})+cos({{9\pi\over13}-{\pi\over13}})+cos{3\pi\over13}+cos{5\pi\over13}\)
\([\because 2cos\ Acos \ B=cos (A+B)+cos(A-B)]\)
\(=cos{10\pi\over 13}+cos{8\pi\over13}+cos{3\pi\over13}+cos{5\pi\over13}\)
\(=cos({\pi-{3\pi\over13}})+cos({\pi-{5\pi\over13}})+cos{3\pi\over13}+cos{5\pi\over13}\)
[\(\because\)cos (\(\pi\) - \(\theta\)) = - cos \(\theta\)]
\(=-cos{3\pi\over13}-cos{5\pi\over13}+cos{3\pi\over13}+cos{5\pi\over13}=0\)
=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) \({ tan15 }^{ 0 }=tan({ 60 }^{ 0 }-{ 45 }^{ 0 })\)
\(=\frac { { tan60 }^{ 0 }-{ tan45 }^{ 0 } }{ 1+{ tan60 }^{ 0 }{ tan45 }^{ 0 } } \left[ tan(A-B)=\frac { tanA-tanB }{ 1+tanAtanB } \right]\)
\(=\frac { \sqrt { 3 } -1 }{ 1+\sqrt { 3 } .1 } \left[ { tan45 }^{ 0 }=1\quad and\quad { tan60 }^{ 0 }=\sqrt { 3 } \right]\)
\(=\frac { \sqrt { 3 } -1 }{ 1+\sqrt { 3 } } =\frac { \sqrt { 3 } -1 }{ \sqrt { 3 } +1 } =\frac { \sqrt { 3 } -1 }{ \sqrt { 3 } +1 } \times \frac { \sqrt { 3 } -1 }{ \sqrt { 3 } -1 } \left[ on\quad rationalisation \right] \)
\(=\frac { { \left( \sqrt { 3 } -1 \right) }^{ 2 } }{ { \sqrt { 3 } }^{ 2 }-\left( { 1 }^{ 2 } \right) } \left[ (a-b)(a+b)={ a }^{ 2 }-{ b }^{ 2 } \right] \)
\(=\frac { { \sqrt { 3 } }^{ 2 }+{ 1 }^{ 2 }-2\sqrt { 3 } }{ 3-1 } \left[ { \left( a-b \right) }^{ 2 }={ a }^{ 2 }{ +b }^{ 2 }-2ab \right] \)
\(=\frac { { \sqrt { 3 } }^{ 2 }+{ 1 }^{ 2 }-2\sqrt { 3 } }{ 3-1 } =\frac { 4-2\sqrt { 3 } }{ 2 } =\frac { 2(2-\sqrt { 3 } ) }{ 2 } =(2-\sqrt { 3 } )\)

cos x + cos y = \(\frac { 1 }{ 3 }\)\(\Rightarrow \) 2cos \(\left( \frac { x+y }{ 2 } \right) \) cos \(\left( \frac { x-y }{ 2 } \right) =\frac { 1 }{ 3 } \) and sin x + sin y = \(\frac { 1 }{ 4 } \)\(\Rightarrow \) 2sin\(\left( \frac { x+y }{ 2 } \right) \) sin \(\left( \frac { x-y }{ 2 } \right) =\frac { 1 }{ 4 } \)

Let x = \(cos\alpha cos\beta \frac { 1 }{ 2 } \)[2\(cos\alpha cos\beta \)]
\(=\frac{1}{2}[\cos (\alpha+\beta)+\cos (\alpha-\beta) \)
\(=\frac{1}{2}\left[\cos 90^{\circ}+\cos (\alpha-\beta)\right]=\frac{1}{2}[\cos ((\alpha-\beta)]\)
since \(-1 \leq \cos (\alpha-\beta) \leq 1-\frac{1}{2} \leq \frac{1}{2} \cos (\alpha-\beta) \leq \frac{1}{2} \Rightarrow-\frac{1}{2} \leq \mathrm{X} \leq \frac{1}{2}\)

\(Let\quad A=\frac { \pi }{ 7 } ,then\)
\(LHS=cos\frac { \pi }{ 7 } cos\frac { 2\pi }{ 7 } cos\frac { { 2 }^{ 2 }\pi }{ 7 } \frac { sin{ 2 }^{ 3 }A }{ { 2 }^{ 3 }sinA }\)
\(=\frac { sin8A }{ 8sinA } =\frac { sin(7A+A) }{ 8sinA } \)
\(=\frac { sin(\pi +A) }{ 8sinA } =\frac { -sinA }{ 8sinA } \quad [\because A=\frac { \pi }{ 7 } ]\)

\(LHS = \left( cos3A+cos7A \right) +(cos5A+cos15A)\)
\(= 2cos5Acos2A+2cos10Acos5A\)
\(= 2cos5A(cos2A+2cos10A)\)
\(= 2cos5A(2cos6Acos4A)\)
\(= 4cos4Acos5Acos6A\)

\(2cos3\theta =2(4cos^{ 3 }\theta -3cos\theta )\)
\(= 8cos^{ 3 }\theta -6cos\theta =(2cos\theta )^{ 3 }-3(2cos\theta )\)
\(= \left( x+\frac { 1 }{ x } \right) ^{ 3 }-3\left( x+\frac { 1 }{ x } \right) \)

\(LHS=(cos^{ 2 }x=sin^{ 2 }x)+(cos^{ 2 }y+sin^{ 2 }y)-2(cosxcosy+sinxsiny)\)
\(=(1+1-2cos(x-y)=2\left[ 1-cos(x-y) \right] \)

\(=\frac { 1 }{ 2 } sin60^{ 0 }sin20^{ 0 }(2sin40^{ 0 }cos80^{ 0 })\)
\(=\frac { 1 }{ 2 } \times \frac { \sqrt { 3 } }{ 2 } sin20^{ 0 }(cos(-40^{ 0 })-cos(120^{ 0 })\)
\(=\frac { \sqrt { 3 } }{ 4 } sin20^{ 0 }\left( cos40^{ 0 }+\frac { 1 }{ 2 } \right) \)
\(=\frac { \sqrt { 3 } }{ 4 } \left[ \frac { 1 }{ 2 } (2sin20^{ 0 }cos40^{ 0 }+\frac { 1 }{ 2 } sin20^{ 0 } \right] \)

LHS=\(\frac { 1 }{ 2 } \) cos60⁰ cos 20⁰(2 cos 40⁰cos 80⁰)
=\(\frac { 1 }{ 2 } \)x\(\frac { 1 }{ 2 } \) cos20⁰(cos120⁰ + cos40⁰)
= \(\frac { 1 }{ 4 } \)\(\left[ \frac { cos20^{ 0 } }{ 2 } +\frac { 1 }{ 2 } (cos60^{ 0 }+cos20^{ 0 } \right] \)
= \(\frac { 1 }{ 4 } \) \(\left[ \frac { 1 }{ 2 } \times \frac { 1 }{ 2 } \right] =\frac { 1 }{ 16 } \)

LHS=tan(360⁰-45⁰)[(-cot(360⁰+45⁰)]+cot(360⁰+135⁰)-[-tan(360⁰+225⁰)]
=-tan45⁰(-cot 45⁰)+cot(135⁰)(-tan 225⁰)
=1-cot(90⁰+45⁰) tan(180⁰+45⁰)

cos 6\(\theta\) + 6 cos 4\(\theta\) + 15 cos 2\(\theta\) + 10
= (cos 6\(\theta\) + cos 4\(\theta\)) + (5 cos 4\(\theta\) + 5 cos 2\(\theta\)) + (10 cos 2\(\theta\) + 10)
= (cos 6\(\theta\) + cos 4\(\theta\)) + 5 (cos 4\(\theta\) + cos 2\(\theta\)) + 10 (cos 2\(\theta\)+ 1)
= 2cos 5\(\theta\) cos \(\theta\) + 5 x 2 cos 3\(\theta\) cos \(\theta\)+ 10 x 2 cos \(\theta\) cos \(\theta\)
= 2cos \(\theta\) [cos5\(\theta\) + 5 cos 3\(\theta\) + 10 cos \(\theta\)]
\(\therefore {cos \theta + 6 cos 4 \theta + 15 cos 2 \theta + 10 \over cos 5 \theta + 5 cos 3 \theta + 10 cos \theta}=2cos \theta\)

Since e lies in third quadrant, sin \(\theta\)is negative and tan\(\theta\) is positive.
\(\therefore sin \theta =-\sqrt{1-cos^2 \theta }\)
\(=-\sqrt{1-({-{1\over 2}})^2}\)
\(=-\sqrt{1-{1\over 4}}=-{\sqrt{3}\over 2}\)
\(tan \theta ={sin \theta \over cos \theta}={-{\sqrt{3}\over2}\over -{1\over2}}=\sqrt{3}\)
\(cosec \theta ={1\over sin \theta}={1\over-{\sqrt{3}\over2}}=-{2\over \sqrt{3}}\)
Now 4 tan² \(\theta\) - 3 cosec²\(\theta\)
\(=4\times (\sqrt{3})^2-3\times (-{2\over \sqrt{3}})^2\)
= 4 x 3- 3 x\(4\over3\)= 12 - 4 = 8.

L.H.S a sin\(\frac { A }{ 2 } \) .sin \(\frac { B-C }{ 2 } \)
= a sin\(\left( \frac { \pi }{ 2 } -\frac { B+C }{ 2 } \right) \).sin \(\left( \frac { B-C }{ 2 } \right) \)
= k sin A. cos \(\left( \frac { B+C }{ 2 } \right) \).sin \(\left( \frac { B-C }{ 2 } \right) \)
= k sin A. [sin B-sin C]
=  [sin A sin B-sin A sin c] ..(i)
Similarly,
b sin \(\frac { B }{ 2 } sin\left( \frac { C-A }{ 2 } \right) \)
=  \(\frac { k }{ 2 } \)[sin B sin C-sin B sin A]... (ii)
and C sin\(\frac { C }{ 2 } \) sin \(\left( \frac { A-B }{ 2 } \right) \)
=\(\frac { k }{ 2 } \) [sin C sin A-sin C sin B] ... (iii)
Adding (i), (ii) and (iii), we get
a sin \(\frac { A }{ 2 } \)sin\(\left( \frac { B-C }{ 2 } \right) \) +b sin\(\frac { B }{ 2 } \)sin \(\left( \frac { C-A }{ 2 } \right) \)+c sin \(\frac { C }{ 2 } \)sin \(\left( \frac { A-B }{ 2 } \right) \)
⇒ \(\frac { k }{ 2 } \) [sin A sin B-sin A sin C] + \(\frac { k }{ 2 } \)[sin B sin C-sin B sin A]+\(\frac { k }{ 2 } \) [sin C sin A-sin C sin B]
⇒ \(\frac { k }{ 2 } \)[sin A sin B-sin A sin C+sin B sin A-sin C sin A-sin C sin B]
⇒ \(\frac { k }{ 2 } \) [0]
⇒  0 R.H.S
Hence proved

Here sin x =  =\({1\over4}\),x in quadrant II.
\(\therefore \ cos^2 x=1-sin^2 x\)
\(\Rightarrow\) cos²x= 1- \(({1\over4})^2=1-{1\over16}={15\over16}\)
\(\therefore cos x=\pm {\sqrt{15}\over4}\)
But x lies in second quadrant.
\(\therefore cos x=- {\sqrt{15}\over4}\)
Also \({x\over2}\)
So \({x\over2}\) lies in first quadrant.
\(\therefore \) sin\(x\over2\) ,cos\(x\over2\)and tan\(x\over2\)are all positive.
Now cos\({x\over2}={\sqrt{1+cos \ x\over 2}}=\sqrt{1+{\sqrt{15}\over4}\over2}\)
\(=\sqrt{4-\sqrt{15}\over 8}={\sqrt{8-2\sqrt{15}\over4}}\)
sin\({x\over2}={\sqrt{1-cos \ x\over 2}}=\sqrt{1+{\sqrt{15}\over4}\over2}\)
\(=\sqrt{4+\sqrt{15}\over 8}={\sqrt{8+2\sqrt{15}\over4}}\)
tan\(x\over2\) \(={sin {x\over2}\over cos {x\over2}}={{\sqrt{8+2\sqrt{15}}\over4}\over{\sqrt{8-2\sqrt{15}}\over4}}\)
\(={\sqrt{4+\sqrt{15}}\over\sqrt{4-\sqrt{15}}}\times {\sqrt{4+\sqrt{15}}\over\sqrt{4+\sqrt{15}}}\)
\(={{4+\sqrt{15}}\over\sqrt{16-16}}=4+\sqrt{15}\)

Here cos x =\({-1\over3},x\) in quadrant III.
Now \(\pi\)\({3\pi\over 2} \Rightarrow {\pi\over 2}<{x\over2}<{3\pi\over 4}\)
So\(x\over2\)  lies in second quadrant.
\(\therefore\) sin \(x\over2\) is positive and cos\(x\over2\),tan\(x\over2\) are negative.
Now cos \({x\over2}=-{\sqrt{1+cos \ x\over 2}}=\sqrt{1-{1\over3}\over2}\)
\(=-\sqrt{1\over\sqrt{3}}\times {\sqrt{5}\over \sqrt{5}}={-\sqrt{3}\over3}\)
sin\({x\over2}={\sqrt{1+cos \ x\over 2}}=\sqrt{1-{1\over3}\over2}\)
\(=\sqrt{2\over3}\times {\sqrt{3}\over\sqrt{3}}={\sqrt{6}\over3}\)
tan \({x\over2}={sin{x\over2}\over cos{x\over2}}={\sqrt{2\over3}\over-\sqrt{1\over \sqrt{3}}}=-\sqrt{2}\)

From Sine formula
Let \(\frac{a}{sinA}=\frac{b}{sinB}=\frac{c}{sinC}=k\)
Therefore, \(\frac{b-c}{b+c}=\frac{ksinB-ksinC}{ksinB+ksinC}\)
\(=\frac{sinB-sinC}{sinB+sinC}=\frac{2cos\frac{B+C}{2}.sin\frac{B-C}{2}}{2sin\frac{B+C}{2}.cos\frac{B-C}{2}}\)
\(=cot\frac{B+C}{2}.tan\frac{B-C}{2}=cot(\frac{\pi}{2}-\frac{A}{2}).tan\frac{B-C}{2}\)
\(=tan\frac{A}{2}.tan\frac{B-C}{2}=\frac{tan\frac{B-C}{2}}{cot\frac{A}{2}}\)
Therefore, \(tan\frac{B-C}{2}=\frac{b-c}{b+c}cot\frac{A}{2}\)
Similarly, we can prove other results also. These are known as Napier's Analogy.

\(1\over8\)

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

We have L.H.S.
= cot 4x (sin 5x + sin 3x)
\(={cos \ 4x\over sin \ 4x}[2sin({5x+3x\over 2})cos ({5x-3x\over2})]\)
\(={cos 4x\over sin 4x}[2sin 4xcosx]\)
=2cos4x cos x
Wehave R.H.S.
= cot x [sin 5x - sin 3x]
\(={cos x\over sin x}[2cos({5x+3x\over 2})sin ({5x-3x\over2})]\)
\([\because \ sin C-sin D=2cos{C+D\over 2}:{C-D\over2}]\)
\(={cos \ x\over sin \ x}[2cos \ 4x \ sin x]\)
= 2 cos 4x cos x
Hence L.H.S. = R.H.S.

-1

\({-2\over3}\)

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

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

\(LHS=tan82\frac { { 1 }^{ \circ } }{ 2 } =tan\left( 90^{ \circ }-7\frac { 1^{ \circ } }{ 2 } \right) =cot7\frac { 1^{ \circ } }{ 2 } =cotA\quad (say)\)
\(where,\quad A=7\frac { 1^{ \circ } }{ 2 } \)
\(Now,\quad cotA=\frac { cosA }{ sinA } =\frac { cosA(2cosA) }{ sinA(2cosA) } =\frac { 1+cos2A }{ sin2A } \)
\(\because\ cot7\frac { 1^{ \circ } }{ 2 } =\frac { 1+cos2\left( \frac { 15 }{ 2 } \right) ^{ \circ } }{ sin2\left( \frac { 15 }{ 2 } \right) ^{ \circ } } =\frac { 1+cos15^{ \circ } }{ sin15^{ \circ } } =\frac { 1+cos(45^{ \circ }-30^{ \circ }) }{ sin(45^{ \circ }-30^{ \circ }) } \)
\(= \frac { 1+\left( \frac { 1 }{ \sqrt { 2 } } .\frac { \sqrt { 3 } }{ 2 } +\frac { 1 }{ \sqrt { 2 } } .\frac { 1 }{ 2 } \right) }{ \left( \frac { 1 }{ \sqrt { 2 } } .\frac { \sqrt { 3 } }{ 2 } -\frac { 1 }{ \sqrt { 2 } } .\frac { 1 }{ 2 } \right) } =\left( \sqrt { 2 } +1 \right) \left( \sqrt { 3 } +\sqrt { 2 } \right) \)

tan\(\theta\)+tan2\(\theta\)=-tan 3\(\theta\)(1-tan\(\theta\)tan2\(\theta\))
\(\Rightarrow\) -tan3\(\theta\)=\(\frac { tan\theta +tan2\theta }{ 1-tan\theta tan2\theta } \)
\(\Rightarrow\)-tan3\(\theta\)=tan3\(\theta\)\(\Rightarrow\)tan3\(\theta\)=0
Ans
\(\theta = \frac { n\pi }{ 3 } ,n\in Z\)

\(cos \theta={4\over5},tan \theta={3\over4},sec\theta={5\over4},cot\theta {4\over3},cosec\theta={5\over 3}\)

6° 21' 49\(1\over 11\)''

\({\pi\over 20},{9\pi\over 20}\)

- 47° 30' = -\((47{30\over 60})^c=-({95\over 2})^o\)
\(=-({95\over 2}\times{\pi\over 180})^c=-({19\pi\over 72})^c\)

\(\operatorname{cosec}=-2 \Rightarrow \sin x=\frac{-1}{2}\)
\(x=\frac{7 \pi}{6}, \frac{11 \pi}{6}\)

\(\tan x=\frac{1}{\sqrt{3}} \Rightarrow \tan x=\tan \frac{\pi}{6}, \tan \frac{7 \pi}{6}\)
\(x=\frac{\pi}{6}, \frac{7 \pi}{6}\)

\(RHS=cos(\theta -\phi )+\cfrac { cos(x+\phi ) }{ sin(x+\phi ) } sin(\theta -\phi )\)
\(=\cfrac { cos(\theta -\phi )sin(x+\phi )+cos(x+\phi )sin(\theta -\phi ) }{ sin(x+\phi ) } \)
\(=\cfrac { sin(x+\phi +\theta -\phi ) }{ sin(x+\phi ) } =\cfrac { sin(x+\theta ) }{ sin(x+\phi ) } \)

(i) (d) Hint \(a=\sqrt{3}\) and b=-2
\(\therefore x\) lies in 4 th quadrant.
\( \cot x=\frac{a}{b}=-\sqrt{3}=\cot \left(2 \pi-\frac{\pi}{6}\right) \)
\( \cot x=\cot \frac{11 \pi}{6} \Rightarrow x=\frac{11 \pi}{6}\)
(ii) (a) Hint \(\tan 75^{\circ} =\tan \left(45^{\circ}+30^{\circ}\right)=\frac{\tan 45^{\circ}+\tan 30^{\circ}}{1-\tan 45^{\circ} \tan 30^{\circ}} \)
\(=\frac{1+\frac{1}{\sqrt{3}}}{1-\frac{1}{\sqrt{3}}}=\frac{\sqrt{3}+1}{\sqrt{3}-1}=2+\sqrt{3}\)
(iii) (c) Hint \(x=25^{\circ}\)
\(25^{\circ}=\left(25 \times \frac{\pi}{180^{\circ}}\right) \text { radian }=\frac{5 \pi}{36}\)
(iv) (d) Hint cos x is negative in \(x \in\left(\frac{\pi}{2}, \frac{3 \pi}{2}\right)\).