Question

Can pls solve it the question is based on trigonometry​

Answer 1

Answer:

Just break tan and cot, rest is simple calculation.

Answer 2

Question:-

Prove the identity, where θ is an acute angle

Identity:-

$\sf{\dfrac{tan\theta}{1-cot\theta} + \dfrac{cot\theta}{1-\tan\theta} = 1 + sec\theta cosec\theta}$

To Prove:-

LHS = RHS

Solution:-

Taking LHS

$\sf{\dfrac{tan\theta}{1-cot\theta} + \dfrac{cot\theta}{1-tan\theta}}$

We know,

$\sf{tan\theta = \dfrac{sin\theta}{cos\theta}}$

$\sf{cot\theta = \dfrac{cos\theta}{sin\theta}}$

Hence,

$\sf{\dfrac{\dfrac{sin\theta}{cos\theta}}{1-\dfrac{cos\theta}{sin\theta}} + \dfrac{\dfrac{cos\theta}{sin\theta}}{1-\dfrac{sin\theta}{cos\theta}}}$

= $\sf{\dfrac{\dfrac{sin\theta}{cos\theta}}{\dfrac{sin\theta - cos\theta}{sin\theta}} + \dfrac{\dfrac{cos\theta}{sin\theta}}{\dfrac{cos\theta - sin\theta}{cos\theta}}}$

= $\sf{\dfrac{sin^2\theta}{cos\theta(sin\theta - cos\theta)} + \dfrac{cos^2\theta}{sin\theta(cos\theta - sin\theta)}}$

= $\sf{\dfrac{sin^2\theta}{cos\theta(sin\theta - cos\theta)} - \dfrac{cos^2\theta}{sin\theta(sin\theta - cos\theta)}}$

Taking LCM,

$\sf{\dfrac{sin^3\theta - cos^3\theta}{cos\theta sin\theta(sin\theta - cos\theta)}}$

Using the identity,

x³ - y³ = (x - y)(x² + xy + y²)

= $\sf{\dfrac{(sin\theta - cos\theta)(sin^2\theta + sin\theta cosec\theta + cos^2\theta)}{cos\theta sin\theta (sin\theta - cos\theta)}}$

= $\sf{\dfrac{sin^2\theta + sin\theta cos\theta + cos^2\theta}{sin\theta cos\theta}}$

= $\sf{\dfrac{sin^2\theta + cos^2\theta + sin\theta cos\theta}{sin\theta cos\theta}}$

We know,

sin²θ + cos²θ = 1

= $\sf{\dfrac{1 + sin\theta cos\theta}{sin\theta cos\theta}}$

= $\sf{\dfrac{1}{sin\theta cos\theta} + \dfrac{sin\theta cos\theta}{sin\theta cos\theta}}$

= $\sf{\dfrac{1}{sin\theta cos\theta} + 1}$

= $\sf{\dfrac{1}{sin\theta}\times\dfrac{1}{cos\theta} + 1}$

We know,

$\sf{\dfrac{1}{sin\theta} = cosec\theta}$

$\sf{\dfrac{1}{cos\theta} = sec\theta}$

Therefore,

= $\sf{cosec\theta sec\theta + 1}$

=> $\sf{1 + sec\theta cosec\theta}$

Hence, LHS = RHS (Proved)

______________________________________