Given, \(\tan { A } =a\tan { B } \Rightarrow \tan { B } =\frac { 1 }{ a } \tan { A } \)
\(\Rightarrow \cot { A } =\frac { a }{ \tan { A } } \quad \quad ..(i)\)
and \(\sin { A } =b\sin { B } \Rightarrow \sin { B } =\frac { 1 }{ b } \sin { A } \)
\(\Rightarrow cosecB=\frac { b }{ \sin { A } } \quad \left[ \because cosec\theta =\frac { 1 }{ \sin { \theta } } \right] \quad ...(ii)\)
We know that, \({ cosec }^{ 2 }B-\cot ^{ 2 }{ B } =1\)
\(\Rightarrow \frac { { b }^{ 2 } }{ \sin ^{ 2 }{ A } } -\frac { { a }^{ 2 } }{ \tan ^{ 2 }{ A } } =1\)  [from Eqs. (i) and (ii)]
\(\Rightarrow \frac { { b }^{ 2 } }{ \sin ^{ 2 }{ A } } -\frac { { a }^{ 2 }\cos ^{ 2 }{ A } }{ \sin ^{ 2 }{ A } } \quad \left[ \because \tan { A } =\frac { \sin { A } }{ \cos { A } } \right] \)
\(\Rightarrow \frac { { b }^{ 2 }-{ a }^{ 2 }\cos ^{ 2 }{ A } }{ \sin ^{ 2 }{ A } } =1\)
\(\Rightarrow { b }^{ 2 }-{ a }^{ 2 }\cos ^{ 2 }{ A } =\sin ^{ 2 }{ A } \)
\(\Rightarrow { b }^{ 2 }-{ a }^{ 2 }\cos ^{ 2 }{ A } =1-\cos ^{ 2 }{ A } \)
\(\Rightarrow \quad { b }^{ 2 }-1={ a }^{ 2 }\cos ^{ 2 }{ A } -\cos ^{ 2 }{ A } \)
\(\Rightarrow \frac { { b }^{ 2 }-1 }{ { a }^{ 2 }-1 } =\cos ^{ 2 }{ A } \)
Hence proved.

LHS=\({ (cosec\theta -\cot { \theta } ) }^{ 2 }={ \left( \frac { 1 }{ \sin { \theta } } -\frac { \cos { \theta } }{ \sin { \theta } } \right) }^{ 2 }\) \(\left[ \because cosec \quad A =\frac { 1 }{ \sin { A } } ,\cot { A } =\frac { \cos { A } }{ \sin { A } } \right] \)
\(={ \left( \frac { 1-\cos { \theta } }{ \sin { \theta } } \right) }^{ 2 }=\frac { { (1-\cos { \theta } ) }^{ 2 } }{ \sin ^{ 2 }{ \theta } } =\frac { { (1-\cos { \theta } ) }^{ 2 } }{ 1-\cos ^{ 2 }{ \theta } } \left[ \because \sin ^{ 2 }{ A } =1-\cos ^{ 2 }{ A} \right] \)
\(=\frac { (1-\cos { \theta } )(1-\cos { \theta } ) }{ (1+\cos { \theta } )(1-\cos { \theta } ) } \left[ \because \quad { a }^{ 2 }-{ b }^{ 2 }=(a+b)(a-b) \right] \)
\(=\frac { 1-\cos { \theta } }{ 1+\cos { \theta } } =RHS\)
Hence proved.

\(\therefore \sec { \theta } +\tan { \theta } =p\)    . ..(i)
LHS = \(\sec { \theta } -\tan { \theta } \times \frac { \sec { \theta } +\tan { \theta } }{ \sec { \theta } +\tan { \theta } } \)
\(=\frac { 1 }{ \sec { \theta } +\tan { \theta } } =\frac { 1 }{ p } \quad \quad \quad ....(i)\)
On adding Eqs. (i) and (ii), we get
\(2\sec { \theta } =p+\frac { 1 }{ p } \Rightarrow \frac { 2p }{ { p }^{ 2 }+1 } \)
\(\therefore \quad \sin { \theta } =\sqrt { 1-{ \left( \frac { 2p }{ { p }^{ 2 }+1 } \right) }^{ 2 } } =\frac { { p }^{ 2 }-1 }{ { p }^{ 2 }+1 } \) 

0

Use Pythagoras theorem, to find hypotenuse.
\(\frac { \sqrt { { a }^{ 2 }+{ b }^{ 2 } } }{ a } \)

Statue=CD = 1.6 m
h = height of pedestal=BC
A is point on earth
In \(\triangle\)ABD, \(\cot { { 60 }^{ 0 } } =\frac { AB }{ BD } \)
\(\Rightarrow \quad \frac { 1 }{ \sqrt { 3 } } =\frac { AB }{ h+1.6 } \)
\(\Rightarrow \quad AB=\frac { h+1.6 }{ \sqrt { 3 } } \quad ...(i)\)
\(\triangle\)ABC, \(\frac { AB }{ BC } =\cot { { 45 }^{ 0 } } \)
\(\Rightarrow \quad \frac { 1 }{ \sqrt { 3 } } =\frac { AB }{ h+1.6 } \)
\(\Rightarrow \quad \frac { 1 }{ \sqrt { 3 } } =\frac { AB }{ h+1.6 } \)
\(\Rightarrow \quad AB=h\)
From (i) and (ii), we get
\(\Rightarrow \quad h=\frac { h+1.6 }{ \sqrt { 3 } } \)
Height of pedestal = h =2.2m.

\(LHS=\left( cosec\theta -\sin { \theta } \right) \left( \sec { \theta } -\cos { \theta } \right) \left( \tan { \theta } +\cot { \theta } \right) \)
\(=\left( \frac { 1 }{ \sin { \theta } } -\sin { \theta } \right) \left( \frac { 1 }{ \cos { \theta } } -\cos { \theta } \right) \left( \frac { \sin { \theta } }{ \cos { \theta } } +\frac { \cos { \theta } }{ \sin { \theta } } \right) \)
\(=\left( \frac { 1-\sin ^{ 2 }{ \theta } }{ \sin { \theta } } \right) \left( \frac { 1-\cos ^{ 2 }{ \theta } }{ \cos { \theta } } \right) \left( \frac { \sin ^{ 2 }{ \theta } +\cos ^{ 2 }{ \theta } }{ \sin { \theta } .\cos { \theta } } \right) \)
\(=\frac { \cos ^{ 2 }{ \theta } }{ \sin { \theta } } \times \frac { \sin ^{ 2 }{ \theta } }{ \cos { \theta } } \times \left( \frac { 1 }{ \sin { \theta } .\cos { \theta } } \right) \)
\(\left[ \because \quad \sin ^{ 2 }{ \theta } +\cos ^{ 2 }{ \theta } =1 \right] \)
= 1
= RHS

Given, a \(\triangle\) ABC, in which B=90⁰
In a \(\triangle\) ABC, \(tanA=\frac { Perpendicular }{ base } \)
\(=\frac { BC }{ AB } =1\)
BC=AB
Let AB=BC=k, where k is a positive number.
Now,  \(AC=\sqrt { { AB }^{ 2 }+{ BC }^{ 2 } } \)       [by unsing Pythagoras theorem]
\(=\sqrt { { (k) }^{ 2 }+{ (k) }^{ 2 } } =k\sqrt { 2 } \)
\(\therefore \quad sinA=\frac { Perpendicular }{ Hypotenuse } =\frac { BC }{ AC } =\frac { k }{ \sqrt { 2 } k } =\frac { 1 }{ \sqrt { 2 } } \)
\(cosA=\frac { Base }{ Hypotenuse } =\frac { AB }{ AC } =\frac { k }{ \sqrt { 2 } k } =\frac { 1 }{ \sqrt { 2 } } \)
Now, LHS=2sin A cos A=\(2\left( \frac { 1 }{ \sqrt { 2 } } \right) \left( \frac { 1 }{ \sqrt { 2 } } \right) =1=RHS\)
Hence proved.

\(\frac { \sin { { 70 }^{ 0 } } }{ \cos { { 20 }^{ 0 } } } +\frac { cosec{ 36 }^{ 0 } }{ \sec { { 54 }^{ 0 } } } -\frac { 2\cos { { 43 }^{ 0 } } cosec{ 47 }^{ 0 } }{ \tan { { 10 }^{ 0 } } \tan { { 40 }^{ 0 } } \tan { { 50 }^{ 0 } } \tan { { 80 }^{ 0 } } } \)
\(=\frac { \sin { { ({ 90 }^{ 0 }-20 }^{ 0 }) } }{ \cos { { 20 }^{ 0 } } } +\frac { cosec{ ({ 90 }^{ 0 }-54 }^{ 0 }) }{ \sec { { 54 }^{ 0 } } } -\frac { 2\cos { { 43 }^{ 0 } } cosec{ ({ 90 }^{ 0 }-43 }^{ 0 }) }{ \tan { { ({ 90 }^{ 0 }-80 }^{ 0 }) } \tan { { ({ 90 }^{ 0 }-50 }^{ 0 }) } \tan { { 50 }^{ 0 } } \tan { { 80 }^{ 0 } } } \)
\(=\frac { \cos { { 20 }^{ 0 } } }{ \cos { { 20 }^{ 0 } } } +\frac { \sec { { 54 }^{ 0 } } }{ \sec { { 54 }^{ 0 } } } -\frac { 2\cos { { 43 }^{ 0 } } \sec { { 43 }^{ 0 } } }{ \cot { { 80 }^{ 0 } } \cot { { 50 }^{ 0 } } \tan { { 50 }^{ 0 } } \tan { { 80 }^{ 0 } } } \)
\(\left[ \because \sin { ({ 90 }^{ 0 }-\theta ) } =\cos { \theta } ,cosec({ 90 }^{ 0 }-\theta )=\sec { \theta } ,\tan { ({ 90 }^{ 0 }-\theta ) } =\cot { \theta } \right] \)
\(=2-\frac { 2.1 }{ 2.1 } =2-2=0\) 

We have, \(\frac { \cos { { 60 }^{ 0 } } +\sin { { 45 }^{ 0 } } -\cot { { 30 }^{ 0 } } }{ \tan { { 60 }^{ 0 } } +\sec { { 45 }^{ 0 } } -cosec{ 30 }^{ 0 } } \)
\(=\frac { \frac { 1 }{ 2 } +\frac { 1 }{ \sqrt { 2 } } -\sqrt { 3 } }{ \sqrt { 3 } +\sqrt { 2 } -2 } =\frac { \frac { \sqrt { 2 } +2-2\sqrt { 2 } \times \sqrt { 3 } }{ 2\sqrt { 2 } } }{ \sqrt { 3 } +\sqrt { 2 } -2 } \)
\([ \because \cos { { 60 }^{ 0 } } =1/2,\sin { { 45 }^{ 0 } } =1/\sqrt { 2 } ,\cot { { 30 }^{ 0 } } =\sqrt { 3 }\)
\( \tan { { 60 }^{ 0 } } =\sqrt { 3 } ,\sec { { 45 }^{ 0 } } =\sqrt { 2 } \quad and\quad cosec{ 30 }^{ 0 }=2 ] \)
\(=\frac { \sqrt { 2 } +2-2\sqrt { 6 } }{ 2\sqrt { 2 } (\sqrt { 3 } +\sqrt { 2 } -2) } \)
\(=\frac { \sqrt { 2 } +2-2\sqrt { 6 } }{ 2\sqrt { 6 } +2\sqrt { 2 } \times \sqrt { 2 } -2\sqrt { 2 } \times 2 } =\frac { \sqrt { 2 } +2-2\sqrt { 6 } }{ 2\sqrt { 6 } +4-4\sqrt { 2 } } \) 

Let the angle of elevation of top of the tower be θ.
From ΔABC,
\({AB\over BC}=tan\theta\)
\(\Rightarrow\ {10\sqrt3\over 30}=tan\theta\)
\(\Rightarrow\ tan\theta={1\over\sqrt3}\)
θ=30° 
Hence angle of elevation is θ.

Let BD and AE be two poles
Where BD = 16 m, AE = 10m
Length BC = BD - CD
= BD-AE
= 16-10 = 6m
From ∆ABC
\({BC\over l}={sin30^0}\)
\({6\over l}={1\over 2}\)
1 = 6 x 2 = 12 m.

\({Length\ of\ rod\over Length\ of\ shadow}={1\over 1/\sqrt3}\)
\({p\over b}=tan\theta={1\over 1/\sqrt3}\)
\(tan\theta=\sqrt3=tan60^0\)
\(\theta=60^0\)

Given, 5cosec \(\theta \) = 7
\(\Rightarrow \quad cosec\theta =\frac { 7 }{ 5 } \)
\(\Rightarrow \quad \sin { \theta } =\frac { 5 }{ 7 } \quad \left[ \because \quad cosec\theta =\frac { 1 }{ \sin { \theta } } \right] \)
\(\sin { \theta } +\cos ^{ 2 }{ \theta } -1=\sin { \theta } -\left( 1-\cos ^{ 2 }{ \theta } \right) \)
\(\left[ \because \sin ^{ 2 }{ \theta } +\cos ^{ 2 }{ \theta } =1 \right] \)
\(=\frac { 5 }{ 7 } -{ \left( \frac { 5 }{ 7 } \right) }^{ 2 }=\frac { 35-25 }{ 49 } =\frac { 10 }{ 49 } \)

In right triangle ABC
\(cos\ 60^0={BC\over AC}\)
\({1\over 2}={2.5\over AC}\)
\(AC=5.0m\)
Length of the ladder is 5.0 m.

Given, \(\sqrt { 2 } \sin { \theta } =1\)
\(\sin { \theta } =\frac { 1 }{ \sqrt { 2 } } =\sin { { 45 }^{ ° } } \)
\(\therefore \quad \theta ={ 45 }^{ ° }\)
Now \(\sec ^{ 2 }{ \theta } -cosec^{ 2 }\theta \) = \(\sec ^{ 2 }{ { 45 }^{ ° } } -cosec^{ 2 }{ 45 }^{ ° }\)
\(={ \left( \sqrt { 2 } \right) }^{ 2 }-{ \left( \sqrt { 2 } \right) }^{ 2 }\)
= 2 - 2
= 0 

Given \(\sec { \theta } .\sin { \theta } =0\)
\(\Rightarrow \quad \frac { \sin { \theta } }{ \cos { \theta } } =0\)
\(\Rightarrow \quad \tan { \theta } =0=\tan { { 0 }^{ ° } } \)
\(\therefore \quad \theta ={ 0 }^{ ° }\)

Given, \(\sin { \alpha =\frac { 1 }{ 2 } } \)
then \(3\sin { \alpha } -4\sin ^{ 3 }{ \alpha } =3\times \frac { 1 }{ 2 } -4\times { \left( \frac { 1 }{ 2 } \right) }^{ 3 }\)
\(=\frac { 3 }{ 2 } -\frac { 4 }{ 8 } =1\)  

\(\frac {1}{3}\)

Given, \(x=a\sec { \theta } and\quad y=b\tan { \theta } \)
\(\Rightarrow \quad \frac { x }{ a } =\sec { \theta } \quad and\quad \frac { y }{ b } =\tan { \theta } \quad \quad ..(i)\)
We know that, \(\quad \sec ^{ 2 }{ \theta } =1+\tan ^{ 2 }{ \theta } \)
\(\therefore \quad { \left( \frac { x }{ a } \right) }^{ 2 }=1+{ \left( \frac { y }{ b } \right) }^{ 2 }\)  [from Eq. (i)]
\(\Rightarrow \quad \frac { { x }^{ 2 } }{ { a }^{ 2 } } -\frac { { y }^{ 2 } }{ { b }^{ 2 } } =1\)