Question

prove that (2-cos²A)(1+2cot²A)=( 2+cot²A)(2- sin²A)​

Answer 1

Answer:

was....go

Step-by-step explanation:

sinθ

1+cosθ

−

1+cosθ

sinθ

=2cotθ , proved.

Step-by-step explanation:

To prove that, \dfrac{1+\cos \theta}{\sin \theta} -\dfrac{\sin \theta}{1+\cos \theta} =2\cot \theta

sinθ

1+cosθ

−

1+cosθ

sinθ

=2cotθ .

L.H.S. = \dfrac{1+\cos \theta}{\sin \theta} -\dfrac{\sin \theta}{1+\cos \theta}

sinθ

1+cosθ

−

1+cosθ

sinθ

To take LCM of denominator part, we get

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

sinθ(1+cosθ)

(1+cosθ)

2

−sin

2

θ

Using the algebraic identity,

(a+b)^{2} =a^{2} +2ab+b^2(a+b)

2

=a

2

+2ab+b

2

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

sinθ(1+cosθ)

1+cos

2

θ+2cosθ−sin

2

θ

Using the trigonometric identity,

\sin^2 A+\cos^2 A=1sin

2

A+cos

2

A=1

⇒ \sin^2 A=1-\cos^2 Asin

2

A=1−cos

2

A

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

sinθ(1+cosθ)

1+cos

2

θ+2cosθ−(1−cos

2

θ)

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

sinθ(1+cosθ)

1+cos

2

θ+2cosθ−1+cos

2

θ

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

sinθ(1+cosθ)

2cos

2

θ+2cosθ

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

sinθ(1+cosθ)

2cosθ(cosθ+1)

=\dfrac{2\cos \theta}{\sin \theta}=

sinθ

2cosθ

= 2\cot \theta=2cotθ

= R.H.S., proved.

Thus, \dfrac{1+\cos \theta}{\sin \theta} -\dfrac{\sin \theta}{1+\cos \theta} =2\cot \theta

sinθ

1+cosθ

−

1+cosθ

sinθ

=2cotθ , proved.