prove that cot2 theta -tan2 theta=cosec2 theta - sec2 theta
Answers
Answer:
Answer:
To Prove: \frac{cot^2\,\theta(sec\,\theta-1)}{(1+sin\,\theta)}=sec^2\,\theta(\frac{1-sin\,\theta}{1+sec\,\theta})
(1+sinθ)
cot
2
θ(secθ−1)
=sec
2
θ(
1+secθ
1−sinθ
)
Consider,
LHS
=\frac{cot^2\,\theta(sec\,\theta-1)}{(1+sin\,\theta)}=
(1+sinθ)
cot
2
θ(secθ−1)
=\frac{cot^2\,\theta(sec\,\theta-1\times\frac{sec\,\theta+1}{sec\,\theta+1})}{(1+sin\,\theta\times\frac{1-sin\,\theta}{1-sin\,\theta})}=
(1+sinθ×
1−sinθ
1−sinθ
)
cot
2
θ(secθ−1×
secθ+1
secθ+1
)
=\frac{cot^2\,\theta(\frac{sec^2\,\theta+sec\,\theta-sec\,\theta-1}{sec\,\theta+1})}{\frac{1-sin\,\theta+sin\,\theta-sin^2\,\theta}{1-sin\,\theta}}=
1−sinθ
1−sinθ+sinθ−sin
2
θ
cot
2
θ(
secθ+1
sec
2
θ+secθ−secθ−1
)
=\frac{cot^2\,\theta(\frac{sec^2\,\theta-1}{sec\,\theta+1})}{\frac{1-sin^2\,\theta}{1-sin\,\theta}}=
1−sinθ
1−sin
2
θ
cot
2
θ(
secθ+1
sec
2
θ−1
)
=\frac{cot^2\,\theta(\frac{tan^2\,\theta}{sec\,\theta+1})}{\frac{cos^2\,\theta}{1-sin\,\theta}}=
1−sinθ
cos
2
θ
cot
2
θ(
secθ+1
tan
2
θ
)
=(\frac{cot^2\,\theta\times tan^2\,\theta}{sec\,\theta+1})(\frac{1-sin\,\theta}{cos^2\,\theta})=(
secθ+1
cot
2
θ×tan
2
θ
)(
cos
2
θ
1−sinθ
)
=\frac{1(1-sin\,\theta)}{(sec\,\theta+1)cos^2\,\theta}=
(secθ+1)cos
2
θ
1(1−sinθ)
=\frac{sec^2\,\theta(1-sin\,\theta)}{(sec\,\theta+1)}=
(secθ+1)
sec
2
θ(1−sinθ)
=RHS=RHS
Hence Proved.