Question

prove that tan squared theta by 1 + tan squared theta + cot square theta by 1 + cot square theta is equal to 1​

Answer 1

Answer:

(The question is wrong).

Step-by-step explanation:

Proof:

L.H.S.

=(tan^2 teta/1 + tan^2 teta) (cot^2 teta/1 + cot^2 teta)

=(tan^2 teta + tan^2 teta )(cot^2 teta + cot^2 teta)

=(2 tan ^2 teta ) (2 cot^2 teta)

[cot^2 teta = 1/tan^2 teta]

= 2 tan ^2 teta × 1/ 2tan^2 teta

= 1

=R.H.S.

Answer 2

Question

Prove that tan²Ø/(1 + tan²Ø) + cot²Ø/(1 + cot²Ø) = 1

To Prove

tan²Ø/(1 + tan²Ø) + cot²Ø/(1 + cot²Ø) = 1

$\rule{200}2$

Proof

Take L.H.S.

⇒ tan²Ø/(1 + tan²Ø) + cot²Ø/(1 + cot²Ø)

Used identity: 1 + tan²Ø = sec²Ø and 1 + cot²Ø = cosec²Ø

⇒ tan²Ø/sec²Ø + cot²Ø/cosec²Ø

Also, tanØ = sinØ/cosØ, secØ = 1/cosØ, cotØ = cosØ/sinØ and cosecØ = 1/sinØ

⇒ $\sf{\frac{ \frac{sin^2 \theta}{cos^2 \theta} }{ \frac{1}{cos^2 \theta} } \: + \: \frac{ \frac{cos^2 \theta}{sin^2 \theta} }{ \frac{1}{sin^2 \theta} }}$

⇒ $\sf{\frac{sin^2 \theta}{cos^2 \theta} \times \frac{cos^2 \theta}{1} \: + \: \frac{cos^2 \theta}{sin^2 \theta} \times \frac{sin^2 \theta}{1}}$

⇒ sin²Ø + cos²Ø

We know that sin²Ø + cos²Ø = 1

⇒ 1

L.H.S. = R.H.S.

Hence, proved.