LHS=\(\frac { \sin { \theta } -2\sin ^{ 3 }{ \theta } }{ 2\cos ^{ 3 }{ \theta } -\cos { \theta } } \)
\(=\frac { \sin { \theta } (1-2\sin ^{ 2 }{ \theta } ) }{ \cos { \theta } (2\cos ^{ 2 }{ \theta } -1) } \)
\(=\tan { \theta } .\frac { [1-2(1-\cos ^{ 2 }{ \theta } )] }{ 2\cos ^{ 2 }{ \theta } -1 } \)
\(=\tan { \theta } .\frac { 2\cos ^{ 2 }{ \theta } -1 }{ 2\cos ^{ 2 }{ \theta } -1 } =\tan { \theta } =RHS\)
Hence proved.

d = 24

a_n = 1 - 4n
\(\Rightarrow\)  a₁ = 1 - 4 x 1 = -3
a₂ = 1 - 4 x 2 = -7
d = a₂ - a₁
= - 7 - ( - 3 ) = - 4
a₂₅ = a + 24d
- 3 = 24 x ( - 4 ) = - 99
Now,    S₂₅ = \(\frac{25}{2}({a}_{1}+{a}_{25})\)
\(=\frac{25}{2}(-3-99)\)
= 25 x ( - 51 ) = - 1275

Let AB is the tower, AB = 15 m, BC = 15 m 
In right \(\Delta\)ABC,  \(\tan { \theta =\frac { AB }{ BC } } \Rightarrow \tan { \theta } =\frac { 15 }{ 15 } \)
\(\Rightarrow\) \(\tan { \theta } =1 \Rightarrow \theta ={ 45 }^{ o }\)

a = 117,
d = 104 - 117 = -13
a₈ = a + 7d = 117 + 7 x (-13) = 26

Given, \(tan\theta +\frac { 1 }{ tan\theta } =2\)
On squaring both sides, we get
\({ \left( tan\theta +\frac { 1 }{ tan\theta } \right) }^{ 2 }={ \left( 2 \right) }^{ 2 }\)
\(\Rightarrow { tan }^{ 2 }\theta +\frac { 1 }{ { tan }^{ 2 }\theta } +2tan\theta \frac { 1 }{ tan\theta } =4\)
\(\Rightarrow { tan }^{ 2 }\theta +\frac { 1 }{ { tan }^{ 2 }\theta } +2=4\)
\(\Rightarrow { tan }^{ 2 }\theta +\frac { 1 }{ { tan }^{ 2 }\theta } =4-2\)
\(\Rightarrow { tan }^{ 2 }\theta +\frac { 1 }{ { tan }^{ 2 }\theta } =2\)
Hence the value of \(\Rightarrow \frac { 1 }{ { cot }^{ 2 }\theta } +{ cot }^{ 2 }\theta =2\) 

LHS = cot A + tan A = \(\frac { \cos { A } }{ \sin { A } } +\frac { \sin { A } }{ \cos { A } } \)   \(\left[ \because \cot { A } ={ \cos { A } }/{ \sin { A } },\quad \tan { A } ={ \sin { A } }/{ \cos { A } } \right] \)
\(=\frac { \cos ^{ 2 }{ A } +\sin ^{ 2 }{ A } }{ \sin { A } \cos { A } } =\frac { 1 }{ \sin { A } \cos { A } } \left[ \because \cos ^{ 2 }{ A } +\sin ^{ 2 }{ A } =1 \right] \)
\(=\frac { 1 }{ \sin { A } } .\frac { 1 }{ \cos { A } } =cosecA\sec { A } \left[ \because cosecA=\frac { 1 }{ \sin { A } } and\sec { A } =\frac { 1 }{ \cos { A } } \quad \right] \)
= RHS.
Hence proved.

Let  \(3\sqrt { 2 } \)  be a rational number. Then, it will be of the form  \(\frac { p }{ q } \) , where p, q are coprime integers and  \(q\neq 0\).
Now, \(\frac { p }{ q } \) = \(3\sqrt { 2 } \)   \(\Rightarrow \quad \frac { p }{ 3q } =\sqrt { 2 } \)
Since, p is an integer and 3q is also an integer \(\left( 3q\neq 0 \right) \).
So, \(\frac { p }{ 3q } \) is a rational number.
\(\Rightarrow \quad \sqrt { 2 } \) is a rational number.
But this contradicts the fact that \(\sqrt { 2 } \) is an irrational number.
Hence, \(3\sqrt { 2 } \) is an irrational number.

Let the first term of an A.P. = a and common difference = d
Then sum of first n terms is
S_n = \({n \over 2}[2a+(n-1)d]\)
\(\Rightarrow\)    S₁₂ = \({12 \over 2}[2a+(12-1)d]\)
= 6 [ 2a + 11d ] = 12a + 66d
Alos,       S₈ = \({8 \over 2}[2a+(8-1)d]\)
= 4 [ 2a + 7d] = 8a + 28d
S₄ = \({4 \over 2}[2a+(4-1)d]\)
= 2 ( 2a + 3d ) = 4a + 6d
Now. L.H.S
S₁₂ = 12a + 66d
RHS
3( S₈ - S₄ ) = 3 [ ( 8a + 28d ) - ( 4a + 6d ) ]
= 3 [ 8a + 28d - 4a - 6d ]
= 3 [ 4a + 22d ]
= 12a + 66d
\(\therefore\)  LHS = RHS Hence proved.

Let AB is statue, BC is pedestal and BC=x m, CD=y m.
In right ΔBCD, ΔBCD, \(\frac { BC }{ CD } \)=tan 45^o
⇒, \(\frac { x }{ y } \)=1 ⇒ x=y     .....(i)
In right ΔACD, \(\frac { AC }{ CD } \)=tan 600
⇒ \(\frac { x+1.46 }{ y } =\sqrt { 3 } \)
⇒ \(\frac { x+1.46 }{ x } \)=1.73    [Using (i)]
⇒ x+1.46=1.73x ⇒ 0.73x=1.46 ⇒ x=2

Let x = 5 + 3\(\sqrt2\)  where x is a irrational number.
ஃ 3\(\sqrt2\) = x - 5 ⇒ \(\sqrt2\) = \(x-5\over3\)      ......(i)
Asx is rational, \(x-5\over3\) is also  rational.
But \(\sqrt2\) is irrational
∴ (i) presents a contradiction
∴ x is not rational, hence 5 + 3\(\sqrt2\) is irrational

Firstly, put tanA and cot A in terms of sin A and cos A. Then simplify it.

LHS  =\(\sqrt { \frac { 1+\cos { A } }{ 1-\cos { A } } \times \frac { 1+\cos { A } }{ 1+\cos { A } } } \)
\(=\frac { 1+\cos { A } }{ 1-\cos ^{ 2 }{ A } } =\frac { 1+\cos { A } }{ \sin { A } } \)   \([\therefore 1-\cos ^{ 2 }{ A } =\sin ^{ 2 }{ A } ]\)
\(=cosecA+\cot { A } \) 

HCF (60, 84, 108) = 12 and LCM = 3780

Given: A building AB 15 m high and tower CD

Angle of elevation ㄥDAE=30^o
Angle of elevation ㄥDBC=45^o
To find: BC and CD
Solution: In right ΔDEA,
\(\frac { DE }{ x } \)=tan 30^o
[∵ AE=BC=x m]
⇒ \(\frac { h-15 }{ x } =\frac { 1 }{ \sqrt { 3 } } \)
⇒ h-15=\(\frac { x }{ \sqrt { 3 } } \)   ....(i)
In right ΔDCB, h/x=tan 450
⇒ x m ......(ii)
Putting the value of h from equation (ii) in equation (i) 
x-15=\(\frac { x }{ \sqrt { 3 } } \)  [from (i)]
⇒ 15=x-\(\frac { x }{ \sqrt { 3 } } \)
⇒ 15=\(\left( 1-\frac { \sqrt { 3 } }{ 3 } \right) \)x
⇒ 15=\(\left( \frac { 3-\sqrt { 3 } }{ 3 } \right) \)x
⇒ \(\frac { 45 }{ 1.268 } \)=x
x=35.49 m
Putting in (ii) h=35.49 m