Question

sin theta by 1 + cos theta + 1 + cos theta by sin theta is equal to 2 cosec theta​

Answer 1

Answer:

Sin theta/1+cos theta +1+cos theta/sin theta = 2 cosec theta is proved

Solution:

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+(1^{2}+\cos ^{2} \theta+2 \times 1 \times \cos \theta)}{\sin \theta(1+\cos \theta)}=

sinθ(1+cosθ)

sin

2

θ+(1

2

+cos

2

θ+2×1×cosθ)

[(a+b)^{2}=a^{2}+b^{2}+2 \times a \times b][(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

[\sin ^{2} \theta+\cos ^{2} \theta=1][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{gathered}\begin{array}{l}{=\frac{2}{\sin \theta}} \\\\ {=2 \times \frac{1}{\sin \theta}}\end{array}\end{gathered}

=

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

Answer 2

Step-by-step explanation:

i hope this helps you cutie