If sin2θcos2θ(1 + tan2θ)(1 + cot2θ) = λ, then find the value of �

1.	If sin2θcos2θ(1 + tan2θ)(1 + cot2θ) = λ, then find the value of λ.
Answer» Given: sin²θ cos²θ (1 + tan²θ)(1 + cot²θ) = λ To find: λ We know that 1 + tan²θ = sec²θ And 1 + cot²θ = cosec²θ ⇒ sin²θ cos²θ (1 + tan²θ) (1 + cot²θ) = sin²θ cos²θ sec²θ cosec²θ Now, ∵ cosecθ = \(\frac{1}{sinθ}\) ⇒ cosec²θ = \(\frac{1}{sin^2θ}\) And ∵ secθ = \(\frac{1}{cosθ}\) ⇒ sec²θ = \(\frac{1}{cos^2θ}\) ⇒ sin²θ cos²θ (1 + tan²θ) (1 + cot²θ) = sin²θ cos²θ sec²θ cosec²θ = sin²θ cos²θ \(\frac{1}{cos^2θ}\)\(\frac{1}{sin^2θ}\)= 1 ⇒ λ = 1

If sin2θcos2θ(1 + tan2θ)(1 + cot2θ) = λ, then find the value of λ.

Answer»

Given: sin²θ cos²θ (1 + tan²θ)(1 + cot²θ) = λ

To find: λ

We know that 1 + tan²θ = sec²θ

And 1 + cot²θ = cosec²θ

⇒ sin²θ cos²θ (1 + tan²θ) (1 + cot²θ)

= sin²θ cos²θ sec²θ cosec²θ

Now,

∵ cosecθ = \(\frac{1}{sinθ}\)

⇒ cosec²θ = \(\frac{1}{sin^2θ}\)

And ∵ secθ = \(\frac{1}{cosθ}\)

⇒ sec²θ = \(\frac{1}{cos^2θ}\)

⇒ sin²θ cos²θ (1 + tan²θ) (1 + cot²θ)

= sin²θ cos²θ sec²θ cosec²θ

= sin²θ cos²θ \(\frac{1}{cos^2θ}\)\(\frac{1}{sin^2θ}\)= 1

⇒ λ = 1