Question

prove that (cosecQ-cotQ)square =1-cosQ/1+cosQ​

Answer 1

Step-by-step explanation:

To Prove :-

(cosecQ - cotQ)² = 1 - cosQ/ 1 + cosQ

Formula Required :-

1) cosecA = 1/sinA

2) cotA = cosA /sinA

3) sin²A + cos²A = 1

Solution :-

Taking L.H.S :-

= (cosecQ - cotQ)²

= $\left(\dfrac{1}{sinQ} - \dfrac{cosQ}{sinQ}\right)^2$

[ ∴ cosecQ = 1/sinQ , cotQ = cosQ/sinQ ]

= $\left(\dfrac{1-cosQ}{sinQ}\right)^2$

= $\dfrac{(1-cosQ)^2}{(sinQ)^2}$

[∴ (a/b)² = a²/b² ]

= $\dfrac{(1 - cosQ)^2}{sin^2Q}$

= $\dfrac{(1-cosQ)^2}{1 - cos^2Q}$

[ ∴ sin²Q + cos²Q = 1

→ sin²Q = 1 - cos²Q ]

= $\dfrac{(1 - cosQ)^2}{1^2 - cos^2Q}$

= $\dfrac{(1 - cosQ)^2}{(1 + cosQ)\times (1 - cosQ)}$

[∴ a² - b² = (a + b)(a - b) ]

= $\dfrac{(1 - cosQ)(1 - cosQ)}{(1 + cosQ)(1 - cosQ)}$

[ ∴ a² = a × a ]

Cancelling the common terms :-

= $\dfrac{1-cosQ}{1+cosQ}$

= R.H.S

Hence Proved.