prove that √sec-1/sec+1=cosec-cot
Answers
Answered by
1
Answer:
Hi ,
Here I used A instead of theta.
LHS = √( secA - 1) / ( secA + 1 )
=√[(secA-1)²/(secA+1)(secA+1)
= √(secA -1 )²/(sec² A - 1 )
= √(secA - 1 )²/( tan² A )
= ( SecA - 1 ) / tanA
= SecA / tanA - 1/tanA
= ( 1/cosA ) / ( sinA/cosA ) - cotA
= 1/sinA - cotA
= CosecA - cotA
= RHS
I hope this helps you.
Answered by
1
Answer:
hey mate here is your answer
Step-by-step explanation:
LHS = √( secA - 1) / ( secA + 1 )
=√[(secA-1)²/(secA+1)(secA+1)
= √(secA -1 )²/(sec² A - 1 )
= √(secA - 1 )²/( tan² A )
= ( SecA - 1 ) / tanA
= SecA / tanA - 1/tanA
= ( 1/cosA ) / ( sinA/cosA ) - cotA
= 1/sinA - cotA
= CosecA - cotA
= RHS
I hope this helps you.
thank you
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