to prove: 1/(sec A-1) + 1/(sec A+1)= 2 cosec A cot A
Answers
Answered by
83
Solution :
LHS = 1/(secA -1) + 1/(secA+1)
= [secA+1+secA-1]/[(secA-1)(secA+1)]
= (2secA)/[sec²A-1]
=(2secA)/tan²A
= (2/cosA)/(sin²A/cos²A)
= ( 2cosA)/sin²A
= 2(1/sinA)(cosA/sinA)
= 2cosecAcotA
= RHS
••••
LHS = 1/(secA -1) + 1/(secA+1)
= [secA+1+secA-1]/[(secA-1)(secA+1)]
= (2secA)/[sec²A-1]
=(2secA)/tan²A
= (2/cosA)/(sin²A/cos²A)
= ( 2cosA)/sin²A
= 2(1/sinA)(cosA/sinA)
= 2cosecAcotA
= RHS
••••
AkankshaTinni:
thank you
Answered by
98
We know that (a + b)(a - b) = a^2 - b^2
We know that sec^2θ - tan^2θ = 1 (or) sec^2θ - 1 = tan^2θ
⇒ 2 cosecA cotA.
Hope it helps!
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