If bcos theta =a then prove that cosec theta +cot theta =whole root of b+a/b-a
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Answered by
7
hyy dear.....
here ur answer is.....
Let €=A.
Then, by the question,
b cosA=a=>b+a/b-a=secA+1/secA-1
(by componendo and dividendo)
Then, b+a/b-a=1+cosA/1-cosA
=>√(b+a/b-a)=cot(A/2)...by formula of submultiples.
we need to prove that (1+cosA)/sinA=√(b+a/b-a)
now by the formula, 1+cosA=2cos²(A/2)
and sinA= 2sin(A/2)cos(A/2).
substituting we get, cot(A/2)=√(b+a/b-a).
HOPE THIS WILL HELP U ✌️✌️
here ur answer is.....
Let €=A.
Then, by the question,
b cosA=a=>b+a/b-a=secA+1/secA-1
(by componendo and dividendo)
Then, b+a/b-a=1+cosA/1-cosA
=>√(b+a/b-a)=cot(A/2)...by formula of submultiples.
we need to prove that (1+cosA)/sinA=√(b+a/b-a)
now by the formula, 1+cosA=2cos²(A/2)
and sinA= 2sin(A/2)cos(A/2).
substituting we get, cot(A/2)=√(b+a/b-a).
HOPE THIS WILL HELP U ✌️✌️
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