Question

Answer the two questions in attachment. Class X trigonometry...

Answer 1

Answers:-

$(1) \: \frac{tan \theta}{1 - cot \theta} + \frac{cot\theta }{1 - tan \theta} = 1 + tan\theta + cot\theta \: \:$

Solution:-

L.H.S.

$= )\: \frac{tan\theta}{1 - cot\theta} + \frac{cot\theta}{1 - tan\theta} = 1 + tan\theta + cot\theta \: \: \\ \\ = ) \frac{tan\theta}{1 - \frac{1}{tan\theta} } + \frac{ \frac{1}{tan\theta} }{1 - tan\theta} \\ \\ = ) \frac{tan\theta}{ \frac{tan\theta - 1}{tan\theta} } + \frac{1}{tan\theta(1 - tan\theta)} \\ \\ = ) \frac{ {tan}^{2} \theta}{tan\theta - 1} - \frac{1}{tan\theta(tan\theta - 1)} \\ \\ = ) \frac{ {tan}^{3}\theta - 1}{tan\theta(tan\theta - 1)} \\ \\ = ) \frac{(tan\theta - 1)( {tan}^{2}\theta + {1}^{2} + tan\theta) }{tan\theta(tan\theta - 1)} \\ \\ = )\frac{( {tan}^{2}\theta + {1}^{2} + tan\theta) }{tan\theta} \\ \\ = ) \frac{ {tan}^{2} \theta}{tan\theta} + \frac{1}{tan\theta} + \frac{tan\theta}{tan\theta} \\ \\ = )tan\theta + cot\theta + 1$

L.H.S. = R.H.S.

Hence Proved!

Identity Used:-

• tanθ = 1/cotθ
• [ a³ - b³ ] = [ (a - b)(a² + b² + ab) ]

(2)

Given:-

• 3 sinθ + 5 cosθ = 5

To Prove:-

• 5 sinθ - 3 cosθ = ± 3

Proof :-

3 sinθ + 5 cosθ = 5

Squaring on both sides,

=) [ 3 sinθ + 5 cosθ ]² = 5²

=) 9 sin²θ + 25 cos²θ + 30 sinθ.cosθ = 25

=) 9 (1 - cos²θ) + 25 (1-sin²θ) + 30 sinθ.cosθ = 25

=) 9 - 9cos²θ + 25 - 25sin²θ + 30 sinθ.cosθ = 25

=) - ( 9cos²θ + 25sin²θ - 30 sinθ.cosθ ) = 25 - 25 - 9

=) - ( 9cos²θ + 25sin²θ - 30 sinθ.cosθ ) =- 9

=) 9cos²θ + 25sin²θ - 30 sinθ.cosθ = 9

=) ( 5 sinθ )² + ( 3 cosθ )² - (2)(5 sinθ)(3cosθ) = 9

=) [ 5sinθ - 3cosθ ]² = 9

=) 5sinθ - 3cosθ = √9

=) 5sinθ - 3cosθ = ±3

Hence Proved!

Identity Used:-

• sin²θ = ( 1 - cos²θ )
• ( a - b)² = a² + b² - 2ab

Answer 2

Q} 2

SOLUTION÷

Given:-

(1)3 sinθ + 5 cosθ = 5

To Prove:-

5 sinθ - 3 cosθ = ± 3✔✔

Proof :-

3 sinθ + 5 cosθ = 5 (Given)

Squaring on both sides,

=) [ 3 sinθ + 5 cosθ ]² = 5²

=) 9 sin²θ + 25 cos²θ + 30 sinθ.cosθ = 25

[( a - b)² = a² + b² - 2ab]

=) 9 (1 - cos²θ) + 25 (1-sin²θ) + 30 sinθ.cosθ = 25 [sin²θ = ( 1 - cos²θ )]

=) 9 - 9cos²θ + 25 - 25sin²θ + 30 sinθ.cosθ = 25

=) - ( 9cos²θ + 25sin²θ - 30 sinθ.cosθ ) = 25 - 25 - 9

=) - ( 9cos²θ + 25sin²θ - 30 sinθ.cosθ ) =- 9

=) 9cos²θ + 25sin²θ - 30 sinθ.cosθ = 9

=) ( 5 sinθ )² + ( 3 cosθ )² - (2)(5 sinθ)(3cosθ) = 9

=) [ 5sinθ - 3cosθ ]² = 9 [( a - b)² = a² + b² - 2ab]

=) 5sinθ - 3cosθ = √9

=) 5sinθ - 3cosθ = ±3

PROVED

I HOPE YOU WILL UNDERSTAND THE ATTACMENT AND THIS ANSWER ALSO .

IT MIGHT CLEAR YOUR DOUBTS.