Question

Prove that : cosec A-cot A=
A
1 - COS A
1 + cos A​

Answer 1

ANSWER:

To Prove:

• cosec A - cot A = √[(1-cos A)/(1+cos A)]

Proof:

$\text{We are given that,}\\\\:\longrightarrow\csc A-\cot A=\sqrt{\dfrac{1-\cos A}{1+\cos A}}\\\\\text{Solving LHS,}\\\\:\implies\csc A-\cot A\\\\\text{We know that,}\\\\:\hookrightarrow\csc\theta=\dfrac{1}{\sin\theta}\:\:\&\:\:\cot\theta=\dfrac{\cos\theta}{\sin\theta}\\\\\text{So,}\\\\:\implies\csc A-\cot A\\\\:\implies\dfrac{1}{\sin A}-\dfrac{\cos A}{\sin A}\\\\:\implies\dfrac{1-\cos A}{\sin A}\\\\\text{Squaring and then square-rooting the simplified LHS,}\\\\:\implies\sqrt{\left(\dfrac{1-\cos A}{\sin A}\right)^2}\\\\:\implies\sqrt{\dfrac{(1-\cos A)^2}{(\sin A)^2}}\\\\:\implies\sqrt{\dfrac{(1-\cos A)^2}{\sin^2A}}$

$\text{We know that,}\\\\:\hookrightarrow\cos^2\theta+\sin^2\theta=1\implies\sin^2\theta=1-\cos^2\theta.\\\\\text{So,}\\\\:\implies\sqrt{\dfrac{(1-\cos A)^2}{\sin^2A}}\\\\:\implies\sqrt{\dfrac{(1-\cos A)^2}{(1-\cos^2A)}}\\\\:\implies\sqrt{\dfrac{(1-\cos A)^2}{(1^2-\cos^2A)}}\\\\\text{We know that,}\\\\:\hookrightarrow a^2-b^2=(a+b)(a-b)\\\\:\implies\sqrt{\dfrac{(1-\cos A)^2}{(1^2-\cos^2A)}}\\\\:\implies\sqrt{\dfrac{(1-\cos A)^2}{(1-\cos A)(1+\cos A)}}\\\\\text{(1 - cosA) gets cut. So,}\\\\\bf{:\implies\sqrt{\dfrac{1-\cos A}{1+\cos A}}=RHS}\\\\\text{\bf{Hence Proved!!!}}$

Formulae Used:

• cosec A = 1/sin A
• cot A = cos A/sin A
• sin^2 A+cos^2 A = 1
• a^2 - b^2 = (a+b)(a-b)

Answer 2

best answer ke liye refer the attachment.

Agar achha lage to like zarur krna.