Since cos x =\(-\frac{3}{5}\), we have sec x = \(-\frac{5}{3}\)
Now sin² x + cos² x = 1, i.e., sin² x = 1 – cos² x
or sin² x \(=1-\frac{9}{25}=\frac{16}{25}\)
Hence sin x \(=\pm \frac{4}{5}\)
Since x lies in third quadrant, sin x is negative. Therefore
sin x = \(-\frac{4}{5}\)
which also gives
cosec x \(=-\frac{5}{4}\)
Further, we have
\(\tan x=\frac{\sin x}{\cos x}=\frac{4}{3} \text { and } \cot x=\frac{\cos x}{\sin x}=\frac{3}{4} \text { . }\)

We have
L.H.S. = cos 6x = 2 cos² 3x - 1 [cos 2\(\theta\) = 2 cos²\(\theta\) - 1]
=2 [4cos³x-3cosx]²-1 [\(\because\)cos 3\(\theta\) = 4 cos³ \(\theta\) - 3 cas \(\theta\)]
= 2 [16cos⁶ x+ 9 cos²x- 24 cos⁴x]-1
= 32 cos⁶ x + 18 cos² x- 48 cos⁴ x-1
= 32 cos⁶x- 48 cos⁴x + 18 cos² x-1
= R.H.S.

We have
L.H.S. =\(2 sin^2{\pi\over6}+cosec^2{7\pi\over6}cos^2{\pi\over3}\)
Now cosec \({7\pi\over 6}= cosec({\pi+{\pi\over6}})\)
= - cosec \({\pi\over6}=-2\)
\(\therefore \) \(2 sin^2{\pi\over6}+cosec^2{7\pi\over6}cos^2{\pi\over3}\)
\(=2\times({1\over 2})^2+(-2)^2\times ({1\over2})^2\)
\(={2\times 1\over 4}+{4\times1\over4}={6\over4}={3\over2}=R.H.S\)

\(LHS=\quad \frac { sin(A-B) }{ cosA\quad cosB } +\frac { sin(B-C) }{ cosB\quad cosC } +\frac { sin(C-A) }{ cosC\quad cosA }\)
\(\frac { cosC\quad sin(A-B)+cosA\quad sin(B-C)+cosB\quad sin(C-A) }{ cosA \quad cosB \quad cosC }\)
\(=\frac { sinA\quad cosB\quad cosC-cosA\quad cosC\quad sinB+cosA\quad cosC\quad sinB-cosA\quad cosB\quad sinC+cosB\quad cosA\quad sinC-cosB\quad cosC\quad sinA }{ cosA\quad cosB\quad cosC }\)
\(=\frac { 0 }{ cosA\quad cosB\quad cosC } \)
\(=0=RHS\)

RHS
\({ x }^{ 2 }+{ y }^{ 2 }=(acos\theta +bsin\theta )^{ 2 }+(asin\theta -bcos\theta )^{ 2 }\)
\(=({ a }^{ 2 }{ cos }^{ 2 }\theta ={ b }^{ 2 }{ sin }^{ 2 }\theta +2abcos\theta sin\theta )+({ a }^{ 2 }{ sin }^{ 2 }\theta +{ b }^{ 2 }{ cos }^{ 2 }\theta -2absin\theta cos\theta )\)
\(={ a }^{ 2 }({ cos }^{ 2 }\theta +{ sin }^{ 2 }\theta )+{ b }^{ 2 }({ sin }^{ 2 }\theta +{ cos }^{ 2 }\theta )\)
\(={ a }^{ 2 }+{ b }^{ 2 }=LHS\quad [\because { cos }^{ 2 }\theta +{ sin }^{ 2 }\theta =1]\)
Hence proved.

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

14 degree

Given, cos \(\left( \theta +\phi \right) =mcos\left( \theta -\phi \right) \)
\(\Rightarrow \frac { cos(\theta +\phi ) }{ cos(\theta -\phi ) } =\frac { m }{ 1 } \Rightarrow \frac { cos(\theta -\phi ) }{ cos(\theta +\phi ) } =\frac { 1 }{ m } \)
using componendo and dividendo rule, we get
\(\Rightarrow \frac { cos(\theta -\phi )-cos(\theta +\phi ) }{ cos(\theta -\phi )+cos(\theta +\phi ) } =\frac { 1-m }{ 1+m } \)
\(\Rightarrow \frac { -2sin\left( \frac { \theta -\phi +\theta +\phi }{ 2 } \right) .sin\left( \frac { \theta -\phi -\theta -\phi }{ 2 } \right) }{ 2cos\left( \frac { \theta -\phi +\theta +\phi }{ 2 } \right) .cos\left( \frac { \theta -\phi -\theta -\phi }{ 2 } \right) } =\frac { 1-m }{ 1+m } \)
\(\Rightarrow \frac { sin\theta .sin\phi }{ cos\theta .cos\phi } =\frac { 1-m }{ 1+m } \quad \left[ sin\left( -\theta \right) =-sin\theta \quad and\quad cos(-\theta )\quad =\quad cos\theta \right] \)
\(\Rightarrow tan\theta .tan\phi =\frac { 1-m }{ 1+m } \)
\(\Rightarrow tan\theta =\left( \frac { 1-m }{ 1+m } \right) cot\phi \)
Hence proved.
 

\(LHS=cos^{ 3 }A+cos^{ 3 }(120^{ \circ }+A)+cos^{ 3 }(240^{ \circ }+A)\)
\(=\left[ \frac { 3 }{ 4 } cosA+\frac { 1 }{ 4 } cos3A \right] +\left[ \frac { 3 }{ 4 } cos(120^{ \circ }+A)+\frac { 1 }{ 4 } cos3(120^{ \circ }+A) \right] +\left[ \frac { 3 }{ 4 } cos(240^{ \circ }+A)+\frac { 1 }{ 4 } cos3(240^{ \circ }+A) \right] \)
\(=\left[ \frac { 3 }{ 4 } cosA+\frac { 1 }{ 4 } cos3A \right] +\left[ \frac { 3 }{ 4 } cos(120^{ \circ }+A)+\frac { 1 }{ 4 } cos(360^{ \circ }+3A) \right] +\left[ \frac { 3 }{ 4 } cos(240^{ \circ }+A)+\frac { 1 }{ 4 } cos(720^{ \circ }+3A) \right] \)
\(=\frac { 3 }{ 4 } \left[ cosA+cos(120^{ \circ }+A)+cos(240^{ \circ }+A) \right] +\frac { 1 }{ 4 } \left[ cos3A+cos(360^{ \circ }+3A)+cos(360^{ \circ }\times 2+3A) \right] \)
\(=\frac { 3 }{ 4 } \left[ cosA+cos(120^{ \circ }+A)+cos(240^{ \circ }+A) \right] +\frac { 1 }{ 4 } \left[ cos3A+cos3A)+cos3A) \right] \)
\(=\frac { 3 }{ 4 } \left[ cosA+cos(120^{ \circ }+A)+cos(240^{ \circ }+A) \right] +\frac { 1 }{ 4 } \times 3(cos3A)\)
\(=\frac { 3 }{ 4 } \left[ cosA-2cosA.\frac { 1 }{ 2 } \right] +\frac { 3 }{ 4 } cos3A\)
\(=\frac { 3 }{ 4 } \left[ cosA-cosA \right] +\frac { 3 }{ 4 } cos3A=\frac { 3 }{ 4 } cos3A\)

We have,(sin x-cos x)-3(sin 2x-cos 2x)+(sin 3x-cos 3x)=0
\(\Rightarrow [cos(\frac { \pi }{ 2 } -x)-cos\quad x]-3[cos(\frac { \pi }{ 2 } -2x)-\quad cos\quad 2x]+[cos(\frac { \pi }{ 2 } -3x)-cos\quad 3x]=0\)
\(\Rightarrow 2sin(\frac { \pi }{ 4 } )[sin(\frac { \pi }{ 4 } -x)+3sin(\frac { \pi }{ 4 } -2x)-sin(\frac { \pi }{ 4 } -3x)]=0\)
\(\Rightarrow sin(x-\frac { \pi }{ 4 } )-3sin(2x-\frac { \pi }{ 4 } )+sin(3x-\frac { \pi }{ 4 } )=0\)
\(\Rightarrow [sin(x-\frac { \pi }{ 4 } )+sin(3x-\frac { \pi }{ 4 } )]-3sin(2x-\frac { \pi }{ 4 } )=0\)
\(\Rightarrow 2sin(2x-\frac { \pi }{ 4 } )cos(-x)-3sin(2x-\frac { \pi }{ 4 } )=0\)
\( \Rightarrow 2sin(2x-\frac { \pi }{ 4 } )cos\quad x-3sin(2x-\frac { \pi }{ 4 } )=0\)
\(\Rightarrow sin(2x-\frac { \pi }{ 4 } )(2cos\quad x-3)=0\)
\(\Rightarrow sin(2x-\frac { \pi }{ 4 } )=0\quad as\quad cos\quad x\quad \neq \frac { 3 }{ 2 } \quad [\because \quad cos\quad x\quad cannot\quad be\quad greater\quad than\quad 1]\)
\( \therefore sin(2x-\frac { \pi }{ 4 } )=sin\quad 0\)
Ans
\(x=\frac { n\pi }{ 2 } +\frac { \pi }{ 8 } ,n\in Z\)

Using the Identity 20(i), we have
\({ L.H.S. }=\cos \left(\frac{\pi}{4}+x\right)+\cos \left(\frac{\pi}{4}-x\right)\)
\(=2 \cos \left(\frac{\frac{\pi}{4}+x+\frac{\pi}{4}-x}{2}\right) \cos \left(\frac{\frac{\pi}{4}+x-\left(\frac{\pi}{4}-x\right)}{2}\right)\)
\(=2 \cos \frac{\pi}{4} \cos x=2 \times \frac{1}{\sqrt{2}} \cos x=\sqrt{2} \cos x=\mathrm{R} . \mathrm{H.S}\)

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
\(L.H.S =tan \ 4x={2tan\ 2x\over 1-tan^2 \ 2x }\)
\([\because tan 2A={2tan A\over 1-tan^2A} ]\)
\(={2.{2tan \ x\over 1-tan ^2x}\over 1-({2tan \ x\over 1-tan^2 \ x})^2}\)
\(={{4 tan x\over 1-tan^2 x}\over {(1-tan^2 \ x)^2-4tan^2 \ x \over (1-tan^2 \ x)^2}}\)
\(={4tan \ x \over 1-tan^2 x} \times {(1-tan^2 x)^2\over 1+tan^4x-2tan ^2-4tan^2 x}={4tan \ x(1-tan^2 \ x)\over 1-6tan^2 \ x+tan ^4 \ x}\)
=R.H.S

We have
L.H.S. =\({sin 5x + sin 3x \over cos 5x + cos 3x }\)
\(={2sin({5x+3x\over 2})cos({5x-3x\over 2})\over 2cos ({5x+3x\over 2})cos({5x-3x\over 2})}\)
\(\left[\begin{array}{l} \because \sin C+\sin D=2 \sin \left(\frac{\mathbf{C}+\mathbf{D}}{2}\right) \cos \left(\frac{\mathbf{C}-\mathbf{D}}{2}\right) \\ \cos C+\cos \mathbf{D}=2 \cos \left(\frac{\mathbf{C}+\mathbf{D}}{2}\right) \cos \left(\frac{\mathbf{C}-\mathbf{D}}{2}\right) \end{array}\right]\)
\(={2 \sin \ x \cos \ x\over 2 \cos \ 4x \cos \ x}=\tan 4x\)
=R.H.S