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hiii frnd!!!
it's khushi .....
your answer :-
refers to the attachment...
hope this helps you!!!!!
it's khushi .....
your answer :-
refers to the attachment...
hope this helps you!!!!!
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Anonymous:
yes
when we open cosec^2 and sec^2
we get tan^2 and cot^2
solve it by using identities of cot^2 and tan^2
yeah correct !!!! :)
ok ...he understood now..
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LHS = (cosθ + secθ)² + (sinθ + cosecθ)²
= cos²θ + sec²θ + 2(cosθ)(secθ) + sin²θ + cosec²θ + 2(sinθ)(cosecθ)
= cos²θ + sin²θ + sec²θ + cosec²θ + 2(cosθ)(1/cosθ) + 2(sinθ)(1/sinθ)
= 1 + (1+tan²θ) + (1+cot²θ) + 2 + 2
= 5 + 1 + tan²θ + 1 + cot²θ
= 7 + tan²θ + cot²θ
= RHS
Hence proved!
Here, used formula :-
Cosec θ = 1/sin θ
Sec θ = 1/cos θ
Sin²θ + cos²θ = 1
sec²θ = 1+tan²θ
cosec²θ = 1+cot²θ
= cos²θ + sec²θ + 2(cosθ)(secθ) + sin²θ + cosec²θ + 2(sinθ)(cosecθ)
= cos²θ + sin²θ + sec²θ + cosec²θ + 2(cosθ)(1/cosθ) + 2(sinθ)(1/sinθ)
= 1 + (1+tan²θ) + (1+cot²θ) + 2 + 2
= 5 + 1 + tan²θ + 1 + cot²θ
= 7 + tan²θ + cot²θ
= RHS
Hence proved!
Here, used formula :-
Cosec θ = 1/sin θ
Sec θ = 1/cos θ
Sin²θ + cos²θ = 1
sec²θ = 1+tan²θ
cosec²θ = 1+cot²θ
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