Question

If cosec0+cot0 =k then prove that cos0=k^2-1√k^2-1​

Answer 1

your answer ........

Answer 2

Let angle theta be A
Given that,
cosecA+cot A=k
(1/sinA)+(cosA/sinA)=k
((1+cosA)/sin A)=k
1+cos A=k sin A
Squaring on both the sides
(1+cos A)^2=k^2*sin^2A
(1+cosA)^2=k^2(1-cos^2A) (Using the identity sin^2A+cos^2A=1)
(1+cosA)^2=k^2(1+cosA)(1-cosA)
(1+cosA)=k^2(1-cosA)
1+cosA=k^2-k^2cosA
k^2*cosA+cosA=k^2-1
cosA(k^2+1)=k^2-1
cosA=(k^2-1)/(k^2+1)
i.e cos theta=(k^2-1)/k^2+1)
Hence proved