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°\(={\sqrt{10-2\sqrt{5}}\over4}\)​​​​

x =\({n\pi\over 3},n \in z\)

6° 21' 49\(1\over 11\)''

\({ a }^{ 2 }+{ b }^{ 2 }=2+2cos(\alpha -\beta )\)
\(and\quad ab=\left[ 2sin\left( \cfrac { \alpha +\beta }{ 2 } \right) cos\left( \cfrac { \alpha -\beta }{ 2 } \right) \right] \left[ 2cos\left( \cfrac { \alpha +\beta }{ 2 } \right) cos\left( \cfrac { \alpha -\beta }{ 2 } \right) \right] \)
\(=\left[ 2sin\left( \left( \cfrac { \alpha +\beta }{ 2 } \right) \right) cos\left( \cfrac { \alpha -\beta }{ 2 } \right) \right] \left[ 2co{ s }^{ 2 }(\cfrac { \alpha -\beta }{ 2 } ) \right]\)
\(=si{ n }^{ 2 }\left( \cfrac { \alpha +\beta }{ 2 } \right) \left[ 2co{ s }^{ 2 }\cfrac { \alpha -\beta }{ 2 } \right] \)
\(=sin(\alpha +\beta )\left[ 2co{ s }^{ 2 }(\cfrac { \alpha -\beta }{ 2 } ) \right] \)
\( for\quad eq.(i)\)
\({ a }^{ 2 }+{ b }^{ 2 }=2+2\left[ 2co{ s }^{ 2 }(\cfrac { \alpha -\beta }{ 2 } )-1 \right] =4co{ s }^{ 2 }\left( \cfrac { \alpha -2 }{ 2 } \right) \)
\(=\cfrac { 2ab }{ sin(\alpha +\beta ) } \)
\(\Rightarrow sin(\alpha +\beta )=\frac { 2ab }{ { a }^{ 2 }+{ b }^{ 2 } } \)

\(sin7\frac { { 1 }^{ \circ } }{ 2 } =sin\frac { { 15 }^{ \circ } }{ 2 } =\sqrt { \frac { 1-cos{ 15 }^{ \circ } }{ 2 } } =\sqrt { \frac { 1-\frac { \sqrt { 3+1 } }{ 2\sqrt { 2 } } }{ 2 } } \)
\(=\quad \sqrt { \frac { 4-\sqrt { 6-\sqrt { 2 } } }{ 2\sqrt { 2 } } } \)

In 60 min,the minute hand of a watch completes one rotation i.e. it rotates through \(360^{0}\)
∴ Angle traced by the minute hand in 1 min  \(=(\cfrac { 360 }{ 60 } { ) }^{ 0 }={ 6 }^{ 0 }\)
∴ Angle traced by the minute hand in 30min
\(=(30\times 6)^{ 0 }=18{ 0 }^{ 0 }=(180\times \cfrac { \pi }{ 180 } { ) }^{ c }={ \pi }^{ c }\)
\(\because \quad \theta =\cfrac { Arv }{ Radius } \Rightarrow \pi =\cfrac { Arc }{ 1.8 }\)
\(\therefore \quad Arc=\cfrac { 22 }{ 7 } \times 1.8=5.66\quad cm\)

cos 2α cos 2β + sin(α−β+α+β) sin[α−β−(α+β)]
= cos 2α cos 2β + sin 2α sin(−2β)
= cos 2α cos 2β − sin 2α sin 2β

Use the formula of cos (A+B).

Here, \(\pi <\theta <\frac { 3\pi }{ 2 } \) , it means \(\quad \theta \)lies in third quadrant.
We have, 
\(sin\theta =-\frac { 3 }{ 5 } the\quad cosec\quad \theta =\frac { 1 }{ sin\theta } =\frac { 1 }{ -\frac { 3 }{ 5 } } =-\frac { 5 }{ 3 } \)
\(Now,\quad { sin }^{ 2 }\theta +{ cos }^{ 2 }\theta =1\quad \Rightarrow \quad \left( -\frac { 3 }{ 5 } \right) ^{ 2 }+{ cos }^{ 2 }\theta =1\)
\(\Rightarrow \quad { cos }^{ 2 }\theta =1-\frac { 9 }{ 25 } =\frac { 16 }{ 25 } \Rightarrow cos\theta =\pm \sqrt { \frac { 16 }{ 25 } } \)
\(\Rightarrow \quad cos\theta =\pm \frac { 4 }{ 5 } \)

Since \(\theta \) lies in third quadrant. So, cos\(\theta \) is negative.
\(\therefore \quad cos\theta =-\frac { 4 }{ 5 } \Rightarrow sec\theta =\frac { 1 }{ cos\theta } =-\frac { 5 }{ 4 }\)
\(tan\theta =\frac { sin\theta }{ cos\theta } =\frac { -\frac { 3 }{ 5 } }{ -\frac { 4 }{ 5 } } =\frac { 3 }{ 5 } \times \frac { 5 }{ 4 } =\frac { 3 }{ 4 } and\quad cot\theta =\frac { 1 }{ tan\theta } =\frac { 1 }{ \frac { 3 }{ 4 } } =\frac { 4 }{ 3 } \)
On putting the value in given expression, we get
\(\frac { sec\theta -cot\theta }{ tan\theta -cosec\theta } =\frac { \left( -\frac { 5 }{ 4 } -\frac { 4 }{ 3 } \right) }{ \left( \frac { 3 }{ 4 } +\frac { 5 }{ 3 } \right) } =\frac { \left( \frac { -15-16 }{ 12 } \right) }{ \left( \frac { 9+20 }{ 12 } \right) } =-\frac { 31 }{ 29 } \)

\(sin^{ 2 }\frac { \pi }{ 6 } +cos^{ 2 }\frac { \pi }{ 6 } +sin^{ 2 }\frac { \pi }{ 4 } =\left( \frac { 1 }{ 2 } \right) ^{ 2 }+\left( \frac { \sqrt { 3 } }{ 2 } \right) ^{ 2 }+\left( \frac { 1 }{ \sqrt { 2 } } \right) ^{ 2 }\)
\(=\frac { 1 }{ 4 } +\frac { 3 }{ 4 } +\frac { 1 }{ 2 } =\frac { 1+3+2 }{ 4 } =\frac { 6 }{ 4 } =\frac { 3 }{ 2 } \)

According to given information, we have the following figure.
In \(\triangle ARQ\), we have
 \(\angle RAQ={ 60 }^{ \circ }and\quad \angle ARQ={ 90 }^{ \circ }\)   
\(\therefore \angle AQR={ 30 }^{ \circ }\)
Now, in \(\triangle AQP\), we have \(\angle PAQ={ 45 }^{ \circ }\) and \( \angle AQP={ 30 }^{ \circ }\) 

Using sine rule in \(\triangle APQ\), we get
   \(\frac { AP }{ \sin { \angle AQP } } =\frac { PQ }{ \sin { \angle PAQ } } \Rightarrow PQ=35\sqrt { 2 } m\).

\(\underset { x\longrightarrow b }{ lim } \cfrac { { x }^{ 3 }-{ b }^{ 3 } }{ x-b } =\underset { x\longrightarrow 1 }{ lim } \cfrac { { x }^{ 4 }-1 }{ x1 } \)

We have,   \(  \frac { 1 }{ a+b } +\frac { 1 }{ b+c } +\frac { 3 }{ a+b+c } \)
\(\Rightarrow \frac { b+c+a+c }{ (a+c)(b+c) } =\frac { 3 }{ a+b+c } \Rightarrow \frac { a+b+2c }{ (a+c)(b+c) } =\frac { 3 }{ a+b+c } \)
\( \Rightarrow (a+b+2c)(a+b+c)=3(a+c)(b+c)\)
\(\Rightarrow { a }^{ 2 }+b^{ 2 }-{ c }^{ 2 }=ab\)
\(\Rightarrow \frac { \quad { a }^{ 2 }+b^{ 2 }-{ c }^{ 2 } }{ 2ab } =\frac { ab }{ 2ab } \)
\(\Rightarrow cosC=\frac { 1 }{ 2 } =cos60°\quad \therefore \angle C=60°\)
 

LHS= \({ cot }^{ 4 }\theta +{ cot }^{ 2 }\theta \)
\(=\left( { cot }^{ 2 }\theta \right) ^{ 2 }+\left( { cot }^{ 2 }\theta \right) =\left( { cosec }^{ 2 }\theta -1 \right) ^{ 2 }+{ cot }^{ 2 }\theta \quad [\because 1+{ cot }^{ 2 }\theta ={ cosec }^{ 2 }\theta ]\)
\(={ cosec }^{ 4 }\theta +1-2{ cosec }^{ 2 }\theta +{ cosec }^{ 2 }\theta -1\)
\(={ cosec }^{ 4 }\theta -{ cosec }^{ 2 }\theta \)
= R.H.S

\(cos\quad A\quad =\quad \sqrt { 1-\frac { 9 }{ 25 } } =\frac { 4 }{ 5 } ,\)
\(\\ sin\quad B=\quad -\sqrt { 1-\frac { 144 }{ 169 } } =-\frac { 5 }{ 13 } \)
\(tan(A-B)\quad =\quad \frac { 16 }{ 63 } \)

Let the angles of the triangle be
(a - d)^o, a^o and (a + d)^o, Then a - d + a + a + d = 180°
\(\Rightarrow 3a=180^o \Rightarrow a=60^o\)
So angles are (60-d)^o, 60° and (60+ d)^o,
Now
\({number \ of \ grades \ in \ the \ least \ angle \over number \ of \ radians \ in \ the \ greatest \ angle}={40\over \pi}\)
\(\therefore {(60-d)\times {10\over 9}\over(60+d)\times {\pi\over 180}}={40\over \pi}\)
\(\Rightarrow\)600-10d= 120+2d
\(\Rightarrow\)12d= 480
\(\Rightarrow\)d=40
Thus angles of the triangle are (60 - 40)°, 60° and (60+ 40)° i.e. 20°, 60° and 100°.

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 \({x\over2}={sin{x\over2}\over cos{x\over2}}={\sqrt{2\over3}\over-\sqrt{1\over \sqrt{3}}}=-\sqrt{2}\)

x=n\(\pi \ or \ x=n\pi\pm {\pi\over3},n \in z\)

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