We know that sin 36°= \(\sqrt{1-cos^236^o}\)
\(=\sqrt{1-({\sqrt{5}+1\over 4})^2}\)
\(={\sqrt{10-2\sqrt{5}}\over4}\)
sin 36°\(={\sqrt{10-2\sqrt{5}}\over4}\)​​​​

We know that cos 2\(\theta\) = 1- 2sin² \(\theta\)
\(\therefore cos 36^o=1-2sin^2 18^o\)
\(=1-2[{\sqrt{5}-1\over 4}]^2\)
\(=1-{2(5+1-2\sqrt{5})\over 16}\)
\(={8-6+2\sqrt{5}\over 8}={\sqrt{5}+1\over4}\)
\(\therefore cos 36^o={\sqrt{5}+1\over4}\)

We know that cos \(\theta=\sqrt{1-sin^2 \theta }\)
\(cos 18^o=\sqrt{1-sin ^218^0}\)
\(=\sqrt{1-({\sqrt{5}-1\over 4})^2}\)
\(=\sqrt{1-{{5+1-2\sqrt{5}}\over 16}}\)
\(={\sqrt{10+2\sqrt{5}}\over4}\)
\(cos 18^o={\sqrt{10+2\sqrt{5}}\over4}\)

Here sin\(\theta\) = n sin (\(\theta\)+ 2\(\alpha\).)
\(\Rightarrow {sin (\theta +2\alpha)\over sin \theta}={1\over n}\)
By componendo and dividendo, we have
\({sin(\theta +2\alpha)+sin \theta\over sin (\theta +2\alpha)-sin \theta}={1+n\over 1-n}\)
\(\Rightarrow{2sin(\theta +\alpha)+cos \alpha\over 2sin \alpha cos (\theta+\alpha)}={1+n\over 1-n}\)
\(\Rightarrow{tan(\theta +\alpha)\over tan\alpha }={1+n\over 1-n}\)
\(\Rightarrow{tan(\theta +\alpha) }={1+n\over 1-n}tan \alpha\)

65° 24' 30"

\(1\over \sqrt{3}\)

Given, \(sin\beta =\frac { 1 }{ \sqrt { 10 } } ,0<\beta <\frac { \pi }{ 2 } \text{and } \tan\alpha =\frac { 1 }{ 7 } \)
\(\Rightarrow cos\beta =+\sqrt { 1-sin^{ 2 }\beta }\) [∵β lies in I quadrant]
\(\Rightarrow cos\beta =\sqrt { 1-sin^{ 2 }\beta } \)
\(\Rightarrow cos\beta =\sqrt { 1-\frac { 1 }{ 10 } } \Rightarrow cos\beta =\frac { 3 }{ \sqrt { 10 } } \)
\(tan\beta =\frac { sin\beta }{ cos\beta } =\frac { \frac { 1 }{ \sqrt { 10 } } }{ \frac { 3 }{ \sqrt { 10 } } } =\frac { 1 }{ 3 } \)
\(\text{Now,} \tan2\beta =\frac { 2tan\beta }{ 1-tan^{ 2 }\beta } \)
\(tan2\beta =\frac { 2\times \frac { 1 }{ 3 } }{ 1-\frac { 1 }{ 9 } } =\frac { 2/3 }{ 8/9 } =\frac { 2 }{ 3 } \times \frac { 9 }{ 8 } =\frac { 3 }{ 4 }\)
\(\because tan(\alpha +2\beta )=\frac { tan\alpha +tan2\beta }{ 1-tan\alpha tan2\beta } =\frac { \frac { 1 }{ 7 } +\frac { 3 }{ 4 } }{ 1-\frac { 1 }{ 7 } \times \frac { 3 }{ 4 } } =\frac { 25 }{ 25 } =1\)
\(\Rightarrow \tan(\alpha +2\beta )=1\)
\(\Rightarrow \tan(\alpha +2\beta )=\tan\frac { \pi }{ 4 } \)
\(\because (\alpha +2\beta )=\frac { \pi }{ 4 } \)
Hence proved.

\(LHS=\frac { sinx }{ cos3x } +\frac { sin3x }{ cos9x } +\frac { sin9x }{ cos27x } \)
\(=\frac { 1 }{ 2 } \left[ \frac { 2sinxcosx }{ cos3xcosx } +\frac { 2sin3xcos3x }{ cos9xcos3x } +\frac { 2sin9xcos9x }{ cos27xcos9x } \right] \)
\(=\frac { 1 }{ 2 } \left[ \frac { sin2x }{ cos3xcosx } +\frac { sin6x }{ cos9xcos3x } +\frac { sin18x }{ cos27xcos9x } \right] \)
\(=\frac { 1 }{ 2 } \left[ \frac { sin(3x-x) }{ cos3xcosx } +\frac { sin(9x-3x) }{ cos9xcos3x } +\frac { sin(27x-9x) }{ cos27xcos9x } \right] \)
\(=\frac { 1 }{ 2 } \left[ \frac { sin3xcosx-cos3xsinx }{ cos3xcosx } +\frac { sin9xcos3x-cos9xsin3x }{ cos9x-cos3x } +\frac { sin27xcos9x-cos27xsin9x }{ cos27xcos9x } \right] \)
\(=\frac { 1 }{ 2 } \left( tan3x-tanx+tan9x-tan3x+tan27x-tan9x \right) \)
\( =\frac { 1 }{ 2 } \left( tan27x-tanx \right) \)
Hence proved.

Given equation is 2cos²x+3sinx=0
\([\because { sin }^{ 2 }x+{ cos }^{ 2 }x=1\Rightarrow { cos }^{ 2 }x=1-{ sin }^{ 2 }x]\)
\(\Rightarrow\)2(1-sin²x)+3sinx=0
\(\Rightarrow\)2sin²x-3sinx-2=0
\(\Rightarrow\)2sin²x-4sinx+sinx-2=0
\(\Rightarrow\)2sinx(sinx-2)+1(sinx-2)=0
\(\Rightarrow\)(2sinx+1)(sinx-2)=0
\(\Rightarrow\)sin x=\(\frac { -1 }{ 2 } \) or sin x = 2
Here, sin x=2 is not possible.   \([\because max(sin\theta =1]\)
\(\because \) sin x = \(\frac { -1 }{ 2 } \)= -sin\(\frac { -\pi }{ 6 } \) 

sin 10⁰ + sin 50⁰ - sin 70⁰
 = sin 10⁰ - (sin 70⁰ - sin 50⁰)
= sin10⁰ - \(\left[ 2cos\left( \frac { 70^{ 0 }+50^{ 0 } }{ 2 } \right) sin\quad \left( \frac { 70^{ 0 }-50^{ 0 } }{ 2 } \right) \right] \)
\(\left[ sin\quad A-sin\quad B\quad =2cos\left( \frac { A+B }{ 2 } \right) sin\quad \left( \frac { A-B }{ 2 } \right) \right] \)
=sin 10⁰ - \(\left[ 2cos\left( \frac { 120^{ 0 } }{ 2 } \right) sin\quad \left( \frac { 20^{ 0 } }{ 2 } \right) \right] \)
 =sin 10⁰ - (2 cos 60⁰ sin 10⁰)
 =sin 10^{0 -} \(\left[ 2.\left( \frac { 1 }{ 2 } \right) sin10^{ 0 } \right] \)  \(\left[ cos\quad 60^{ 0 }\frac { 1 }{ 2 } \right] \)
= sin 10⁰ - sin 10⁰ = 0

\(LHS=\frac { { cos15 }^{ 0 }+{ sin15 }^{ 0 } }{ { cos15 }^{ 0 }-{ sin15 }^{ 0 } } =\frac { \frac { { cos15 }^{ 0 } }{ { cos15 }^{ 0 } } +\frac { { sin15 }^{ 0 } }{ { cos15 }^{ 0 } } }{ \frac { { cos15 }^{ 0 } }{ { cos15 }^{ 0 } } -\frac { { sin15 }^{ 0 } }{ { cos15 }^{ 0 } } } \)
\(=\frac { 1+{ tan15 }^{ 0 } }{ 1-{ tan15 }^{ 0 } } \quad \left[ tan\theta =\frac { sin\theta }{ cos\theta } \right]\)
\(=\frac { { tan45 }^{ 0 }+{ tan15 }^{ 0 } }{ 1-{ tan45 }^{ 0 }{ tan15 }^{ 0 } } \left[ { tan45 }^{ 0 }=1 \right] \)
\(=tan({ 45 }^{ 0 }+{ 15 }^{ 0 })\left[ tan(A+B)=\frac { tanA+tanB }{ 1-tanAtanB } \right] \)
\(={ tan60 }^{ 0 }=\sqrt { 3 } =RHS\)

LHS \(=\frac { tan\theta +sec\theta -1 }{ tan\theta -sec\theta +1 }\)
\(=\frac { \left( tan\theta +sec\theta \right) -\left( { sec }^{ 2 }\theta -{ tan }^{ 2 }\theta \right) }{ tan\theta -sec\theta +1 } \) \( [\because { sec }^{ 2 }\theta -{ tan }^{ 2 }\theta =1]\)
\(=\frac { \left( tan\theta +sec\theta \right) \{ 1-(sec\theta -tan\theta )\} }{ tan\theta -sec\theta +1 } \quad [\because { a }^{ 2 }-{ b }^{ 2 }=\left( a-b \right) \left( a+b \right) ]\)
\(=\frac { \left( tan\theta +sec\theta \right) (tan\theta -sec\theta +1) }{ tan\theta -sec\theta +1 } =tan\theta +sec\theta\)
\(=\frac { sin\theta }{ cos\theta } +\frac { 1 }{ cos\theta } =\frac { sin\theta +1 }{ cos\theta } =RHS\)
Hence Proved.

Given cosec \(A=\frac { x }{ y } \)
We known that, cot²A = cosec² A-1
\({ cot }^{ 2 }A=\left( \frac { x }{ y } \right) ^{ 2 }-1=\frac { { x }^{ 2 } }{ { y }^{ 2 } } -1=\frac { { x }^{ 2 }-{ y }^{ 2 } }{ { y }^{ 2 } }\)
\(\Rightarrow \quad cot\quad A=\pm \frac { \sqrt { { x }^{ 2 }-{ y }^{ 2 } } }{ y } \)

Here, given degree measure have minutes.
Firstly, convert minutes into degree.
\(\because \)   \(1'=\left( \frac { 1 }{ 60 } \right) ^{ o }\)
\(\therefore \) \(20'=\left( \frac { 20 }{ 60 } \right) ^{ o }=\left( \frac { 1 }{ 3 } \right) ^{ o }\)
Now, total degree to convert \(={ 40 }^{ o }+\frac { { 1 }^{ o } }{ 3 } \)
\(=\left( \frac { 120+1 }{ 3 } \right) =\left( \frac { 121 }{ 3 } \right) ^{ o }\)
\(\because \) Radian measure = \(\frac { \pi }{ 180 } \times \) Degree measure
\(\therefore \) Required radian measure \(=\frac { \pi }{ 180 } \times \frac { 121 }{ 3 } \)
\(=\frac { 121\pi }{ 540 } radian\)

Required radian measure \(=\frac { \pi }{ 180 } \times \)  Degree measure
\(=\frac { \pi }{ 180 } \times 520rad=\frac { 26 }{ 9 } \pi \)