find sintheta/1+costheta +1+costheta/sintheta
Answers
Answer:
Proof: In the left hand side of the equation it is given that
=\frac{\sin \theta}{1+\cos \theta}+\frac{1+\cos \theta}{\sin \theta}=
1+cosθ
sinθ
+
sinθ
1+cosθ
=\frac{\sin \theta \times \sin \theta+[(1+\cos \theta) \times(1+\cos \theta)]}{\sin \theta(1+\cos \theta)}=
sinθ(1+cosθ)
sinθ×sinθ+[(1+cosθ)×(1+cosθ)]
[Taking L.C.M of sinθ and 1+cosθ]
=\frac{\sin ^{2} \theta+(1+\cos \theta)^{2}}{\sin \theta(1+\cos \theta)}=
sinθ(1+cosθ)
sin
2
θ+(1+cosθ)
2
=\frac{\sin ^{2} \theta+\left(1^{2}+\cos ^{2} \theta+2 \times 1 \times \cos \theta\right)}{\sin \theta(1+\cos \theta)}=
sinθ(1+cosθ)
sin
2
θ+(1
2
+cos
2
θ+2×1×cosθ)
\left[(a+b)^{2}=a^{2}+b^{2}+2 \times a \times b\right][(a+b)
2
=a
2
+b
2
+2×a×b]
=\frac{\sin ^{2} \theta+1+\cos ^{2} \theta+2 \cos \theta}{\sin \theta(1+\cos \theta)}=
sinθ(1+cosθ)
sin
2
θ+1+cos
2
θ+2cosθ
=\frac{1+1+2 \cos \theta}{\sin \theta(1+\cos \theta)}=
sinθ(1+cosθ)
1+1+2cosθ
\left[\sin ^{2} \theta+\cos ^{2} \theta=1\right][sin
2
θ+cos
2
θ=1]
=\frac{2+2 \cos \theta}{\sin \theta(1+\cos \theta)}=
sinθ(1+cosθ)
2+2cosθ
=\frac{2(1+\cos \theta)}{\sin \theta(1+\cos \theta)}=
sinθ(1+cosθ)
2(1+cosθ)
[Taking 2 common]
\begin{lgathered}\begin{array}{l}{=\frac{2}{\sin \theta}} \\\\ {=2 \times \frac{1}{\sin \theta}}\end{array}\end{lgathered}
=
sinθ
2
=2×
sinθ
1
= 2cosecθ = Right Hand Side of the equation.
Thus Sin theta/1+cos theta +1+cos theta/sin theta = 2 cosec theta is proved