Question

(1) (cosec -cot )2 =1-cos/
1 + cos e​

Answer 1

Answer:

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Step-by-step explanation:

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Answer 2

Step-by-step explanation:

$(\csc x-\cot x)^2=\dfrac{1-\cos x}{1+\cos x}\\\\LHS=\left(\dfrac{1}{\sin x}-\dfrac{\cos x}{\sin x}\right)^2=\left(\dfrac{1-\cos x}{\sin x}\right)^2=\dfrac{(1-\cos x)^2}{\sin^2x}\\\\RHS=\dfrac{1-\cos x}{1+\cos x}\cdot\dfrac{1-\cos x}{1-\cos x}=\dfrac{(1-\cos x)^2}{1^2-\cos^2x}=\dfrac{(1-\cos x)^2}{1-\cos^2x}=\dfrac{(1-\cos x)^2}{\sin^2x}\\\\LHS=RHS$

Used:

$\csc\theta=\dfrac{1}{\sin\theta}\\\\\cot\theta=\dfrac{\cos\theta}{\sin\theta}\\\\(a+b)(a-b)=a^2-b^2\\\\\sin^2\theta+\cos^2\theta=1\to\sin^2\theta=1-\cos^2\theta$