\(Cos\quad A=\sqrt { 1-si{ n }^{ 2 } } A\)
\(=\sqrt { 1-\cfrac { 9 }{ 25 } =\sqrt { \cfrac { 16 }{ 25 } =\cfrac { 4 }{ 5 } } } \)
\(\Rightarrow tanA=\cfrac { 3 }{ 4 } \)
\(Also,sinB=-\sqrt { 1-co{ s }^{ 2 } } B\)
\(=-\sqrt { 1-\cfrac { 144 }{ 169 } =-\sqrt { \cfrac { 25 }{ 169 } } =-\cfrac { 5 }{ 13 } } \)
\(=tan B=\cfrac { 5 }{ 12 } \)
\(Now, tan(A-B)=\cfrac { tanA-tanB }{ 1+tanAtanB } =\cfrac { 16 }{ 63 } \)
\(and\quad cot(A+B)=1\)
\(\Rightarrow \frac { cotAcotB-1 }{ cotB+cotA } =1\)
\(\Rightarrow cotAcotB-=cotA+cotB\)
\(\Rightarrow cotAcotB-cot\quad A-cotB=1\)
\( \Rightarrow cotAcotB-cotA-cotB+1=2\)
\(\Rightarrow cotA(cotB-1)-(cotB-1)=2\)
\(\Rightarrow (cotB-1)(cotA-1)=2\)

\(\frac { 3 }{ 4 } \)

given, \(\frac { sin(x+y) }{ sin(x-y) } =\frac { a+b }{ a-b } \)
using componendo and dividendo rule,we get
\(\frac { sin(x+y)+sin(x-y) }{ sin(x+y)-sin(x-y) } =\frac { a+b+a-b }{ a+b-a-b }\)
\( \\ \Rightarrow \frac { 2sin\left( \frac { x+y+x-y }{ 2 } \right) .cos\left( \frac { x+y-x+y }{ 2 } \right) }{ 2cos\left( \frac { x+y+x-y }{ 2 } \right) .sin\left( \frac { x+y-x+y }{ 2 } \right) } =\frac { 2a }{ 2b } \)
\(\left[ sinA+sinB=2sin\frac { A+B }{ 2 } .cos\frac { A-B }{ 2 } and\quad sinA-sinB=2cos\frac { A+B }{ 2 } .sin\frac { A-b }{ 2 } \right] \)
\(\Rightarrow \frac { sin\quad x.cos\quad y }{ cos\quad x.sin\quad y } =\frac { a }{ b } \Rightarrow \quad \frac { tan\quad x }{ tan\quad y } =\frac { a }{ b }\)
Hence proved.

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).

\(LHS\quad =\quad sin\quad (n+1)x\quad sin\quad (n+2)x\quad +\quad cos\quad (n+1)x.cos\quad (n+2)x\)
\(= cos\quad (n+1)x.cos(n+2)x+sin(n+1)x.sin(n+2)x\)
\(= cos\quad [(n+1)x-(n+2)x]\)

LHS = \(\frac { sin\quad 5x+sin3x }{ cos\quad 5x+cos\quad 3x } \)
\(=\frac { 2sin\left( \frac { 5x+3x }{ 2 } \right) cos\left( \frac { 5x-3x }{ 2 } \right) }{ 2scos\left( \frac { 5x+3x }{ 2 } \right) cos\left( \frac { 5x-3x }{ 2 } \right) }\)  [by formulae]
\(=\frac { sin4x\quad cosx }{ cos4x\quad cosx } \)
= tan4x = RHS
Hence proved

\(LHS=\sqrt { 2+\sqrt { 2+\sqrt { 2+2cos8\theta } } }\)
\(= \sqrt { 2+\sqrt { 2+\sqrt { 2(1+cos8\theta ) } } }\)
\(= \sqrt { 2+\sqrt { 2+\sqrt { 2(1+2cos^{ 2 }4\theta -1) } } } \)
\(= \sqrt { 2+\sqrt { 2+\sqrt { 4cos^{ 2 }4\theta } } } =\sqrt { 2+\sqrt { 2+2cos4\theta } } \)
\(= \sqrt { 2+\sqrt { 2(1+cos4\theta ) } } =\sqrt { 2+\sqrt { 2(1+2cos^{ 2 }2\theta -1) } } \)
\(= \sqrt { 2+\sqrt { 4cos^{ 2 }2\theta } } =\sqrt { 2+2cos2\theta } \)
\(= \sqrt { 2(1+cos2\theta } =\sqrt { 2(1+2cos^{ 2 }\theta -1) } \)
\(= \sqrt { 4cos^{ 2 }\theta } =\quad 2cos\theta \)
Hence Proved.

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

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

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
   \(\frac { AP }{ \sin { \angle AQP } } =\frac { PQ }{ \sin { \angle PAQ } } \Rightarrow PQ=35\sqrt { 2 } m\).

\(\underset { x\longrightarrow 1 }{ lim } \cfrac { (x-1)(x+1)({ x }^{ 2 }-2) }{ (x-1)({ x }^{ 2 }-4x-1) } =\cfrac { 1 }{ 2 } \)

\(LHS=\frac { 1 }{ 2 } \left[ 2co{ s }^{ 2 }A+2co{ s }^{ 2 }(A+\frac { 2\pi }{ 3 } )+2co{ s }^{ 2 }(A-\frac { 2\pi }{ 3 } ) \right] \)
\(=\frac { 1 }{ 2 } \left[ 1-cos\quad 2A+\{ 1+co{ s }^{ 2 }(A+\frac { 2\pi }{ 3 } )+\{ 1+co{ s }^{ 2 }(A-\frac { 2\pi }{ 3 } \right] \)
\(=\frac { 1 }{ 2 } \left[ 1+cos\quad 2A+\{ 1+co{ s }(2A+\frac { 4\pi }{ 3 } )+cos(2A-\frac { 4\pi }{ 3 } ) \right] \)
\( =\frac { 1 }{ 2 } \left[ 3+cos2A=2cos2Acos\frac { 4\pi }{ 3 } \right] \)
\(=\frac { 1 }{ 2 } \left[ 3+cos2A+2cos2a(\frac { -1 }{ 2 } ) \right] \)
\(=\frac { 3 }{ 2 } =RHS\)
 

\(We\quad have,\quad { sin }^{ 2 }\theta =1-{ cos }^{ 2 }\theta =1-(-\frac { 3 }{ 5 } )^{ 2 }=\frac { 16 }{ 25 } \)
\(\Rightarrow sin\theta =\pm \frac { 4 }{ 5 } ;\quad but\quad sin\theta \quad is\quad -ve\quad in\quad third\quad quadrant.\)
\(\therefore sin\theta =-\frac { 4 }{ 5 } ;\quad sec\theta =\frac { 1 }{ cos\theta } =\frac { -5 }{ 3 } ;\quad cosec\theta =\frac { 1 }{ sin\theta } =\frac { -5 }{ 4 } \)
\(tan\theta =\frac { sin\theta }{ cos\theta } =\frac { \frac { -4 }{ 5 } }{ \frac { -3 }{ 5 } } =\frac { 4 }{ 3 } \quad and\quad cot\theta =\frac { 1 }{ tan\theta } =\frac { 1 }{ (\frac { 4 }{ 3 } ) } =\frac { 3 }{ 4 }\)
\(Now,\quad \frac { cosec\theta +cot\theta }{ sec\theta -tan\theta } =\frac { -\frac { 5 }{ 4 } +\frac { 3 }{ 4 } }{ -\frac { 5 }{ 3 } -\frac { 4 }{ 3 } } =\frac { \frac { -2 }{ 4 } }{ \frac { -9 }{ 3 } } =\frac { 1 }{ 6 } \)

We have, \(cos \left( \theta +\phi \right) = m\quad cos\left( \theta -\phi \right) \)
\(\Rightarrow \frac { 1 }{ m } =\frac { cos\left( \theta -\phi \right) }{ cos\left( \theta +\phi \right) } \)
Now, \(\frac { 1-m }{ 1+m } =\frac { cos\quad \left( \theta -\phi \right) -cos\quad \left( \theta +\phi \right) }{ cos\quad \left( \theta -\phi \right) +cos\left( \theta +\phi \right) } =tan\quad \theta \quad tan\quad \phi \\ \\ \)
Ans. tan \(\theta \).