1 Answer

0 like 0 dislike
by
selected by
 
Best answer
Left Hand Side

$=\frac{1}{cosecA-1}-\frac{1}{cosecA+1}$

$=\frac{(cosecA+1)-(cosecA-1)}{(cosecA-1)(cosecA+1)}$

$=\frac{cosecA+1-cosecA+1}{cosec^2A-1^2}$

$=\frac{2}{cosec^2A-1}$

$=\frac{2}{cot^2A}$

$=2 \cdot \frac{1}{cot^2A}$

$=2 \cdot tan^2A$

$=$ Right Hand Side
by

Rules Applied :

  • $a^2-b^2=(a+b)(a-b)$
  • $cot^2\theta=cosec^2\theta-1$
  • $tan \theta = \frac{1}{cot \theta}$

Related questions

10.4k questions

10.5k answers

122 comments

27 users

Welcome to QnAfy !

Ask questions, find answers, and spread the light of knowledge like the sun. On QnAfy, only registered users can post questions and answers.

If you are a teacher or student, you may register using your own name, your school/coaching center’s name, or your website’s name to actively contribute by asking or answering questions. This will help increase your or your institution’s visibility, and in the case of a website, it will boost your backlink profile as well.

So, Register Now

fb Group | fb Page
WhatsApp Message
Join Telegram Group

QnAfy – Where curiosity meets clarity.

Categories

For the best experience with math, please rotate your mobile to landscape mode or use a tablet, laptop, or PC for optimal viewing.
...