What should come in the place of (?) for the expression given below?

\(\sqrt{\text {sec}^4 \theta - \text {tan}^4 \theta - \text {tan}^2 \theta}\) × \(\sqrt{\text {cosec}^4 \theta - \text {cot}^4 \theta - \text {cot}^2 \theta}\) = ?

Given:

\(\sqrt{\text {sec}^4 θ - \text {tan}^4 θ - \text {tan}^2 θ}\) × \(\sqrt{\text {cosec}^4 θ - \text {cot}^4 θ - \text {cot}^2 θ}\) = ?

Formula used:

(1) a⁴ - b⁴ = (a² - b²)(a² + b²)

(2) sec2θ - tan2θ = 1

(3) cosec2θ - cot2θ = 1

(4) √a × √b = √(ab)

Calculation:

The above equation can be written as:

⇒ \(\sqrt{(\text {sec}^4 θ - \text {tan}^4 θ - \text {tan}^2 θ) × (\text {cosec}^4 θ - \text {cot}^4 θ - \text {cot}^2 θ)}\)  = ?

Take square on both sides of the equation.

⇒ (sec⁴θ - tan4θ - tan2θ) × (cosec4θ - cot4θ - cot2θ) = ?²

⇒ [(sec4θ - tan4θ) - tan2θ] × [(cosec4θ - cot4θ) - cot2θ] = ?2

⇒ [(sec2θ - tan2θ)(sec2θ + tan²θ) - tan2θ] × [(cosec2θ - cot2θ)(cosec²θ + cot2θ) - cot2θ)] = ?2

⇒ [(1)(sec2θ + tan2θ) - tan2θ] × [(1)(cosec2θ + cot2θ) - cot2θ)] = ?2

⇒ [sec2θ + tan2θ - tan2θ] × [cosec2θ + cot2θ - cot2θ)] = ?2

⇒ [sec2θ] × [cosec2θ] = ?2

Take square root on both side of the equation.

⇒ √[sec2θ × cosec2θ] = ?

⇒ ? = secθ × cosecθ

∴ The required answer is secθ × cosecθ.Additional Information

Most frequently used formulas in exams.

Formula List I:

(1) sin2θ + cos2θ = 1

(2) sec2θ - tan2θ = 1

(3) cosec2θ - cot2θ = 1

Formula List II:

(1) sin4θ - cos4θ = (sin2θ - cos2θ)(sin2θ + cos2θ) = (sin2θ - cos2θ)

(2) sec4θ - tan4θ = (sec2θ - tan2θ)(sec2θ + tan2θ) = (sec2θ + tan2θ)

(3) cosec4θ - cot4θ = (cosec2θ - cot2θ)(cosec2θ + cot2θ) = (cosec2θ + cot2θ)