Question

Prove that


$\sf (cosec \theta - cot \theta)^2 = \dfrac{1- cos \theta}{1+cos \theta}$​

Answer 1

$\:\:\:\: \large\underline {\sf{To\:Prove\::}}$

$\sf (cosec \theta - cot \theta)^2 = \dfrac{1- cos \theta}{1+cos \theta}$

$\:\:\:\: \large\underline {\sf{Need\:To\:Know\::}}$

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

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

$\:\:\:\: \large\underline {\sf{Proof\::}}$

$\sf LHS=(cosec \theta - cot \theta)^2$

$\sf:\implies\left( \dfrac{1}{sin \theta} - \dfrac{cos\theta}{sin \theta} \right)^2$

$:\implies\sf \dfrac{(1-cos \theta)^2}{sin^2 \theta}$

$\bf \dag \ sin^2 \theta = 1- cos^2 \theta$

$\sf:\implies\dfrac{(1-cos \theta)^2}{1-cos^2 \theta}$

$\sf :\implies \dfrac{(1-cos\theta)(1-cos\theta)}{(1+cos\theta)(1-cos\theta)}$

$\bf :\implies \dfrac{1-cos\theta}{1+cos\theta} =RHS$

Proved!

★ Trigonometric Identities :

$\boxed{\begin{minipage}{6cm} Important Trigonometric identities :- \\ \\ \: \: 1)\:\sin^2\theta+\cos^2\theta=1 \\ \\ 2)\:\sin^2\theta= 1-\cos^2\theta \\ \\ 3)\:\cos^2\theta=1-\sin^2\theta \\ \\ 4)\:1+\cot^2\theta=\text{cosec}^2 \, \theta \\ \\5)\: \text{cosec}^2 \, \theta-\cot^2\theta =1 \\ \\ 6)\:\text{cosec}^2 \, \theta= 1+\cot^2\theta \\\ \\ 7)\:\sec^2\theta=1+\tan^2\theta \\ \\ 8)\:\sec^2\theta-\tan^2\theta=1 \\ \\ 9)\:\tan^2\theta=\sec^2\theta-1 \end{minipage}}$

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Answer 2

Answer:

1-cosA/1+cosA

Rationalize it

1-cosA/1+cosA*1-cosA/1-cosA

(1-cosA)^2/(1)^2-cos^2A

(1)^2+cos^2A-2cosA/1-cos^2A

1+cos^2A-2cosA/sin^2A

Apart all individual term between arthimatical operations

1/sin^2A+cos^2A/sin^2A-2cosA/sin^2A

cosec^2A+cot^2A-2cosecA*cotA

(cosecA-cotA)^2

So, RHS=LHS

Formula used

1) a^2-b^2=(a+b)(a-b)

2) (a-b)^2=(a^2+b^2-2ab)

3) cosecA=1/sinA

4) cotA=cosA/sinA

Identity used

1) 1-cos^2A=sin^2A

Step-by-step explanation:

Hope that will help you.