If \(sec\theta = \frac{a}{b},b \ne 0,then\frac{{1 - {{\tan }^2}\theta

1.	If \(sec\theta = \frac{a}{b},b \ne 0,then\frac{{1 - {{\tan }^2}\theta }}{{2 - {{\sin }^2}\theta }}\)=?1. \(\frac{{{b^2}\left( {2{b^2} - {a^2}} \right)}}{{{a^2}\left( {{a^2} + {b^2}} \right)}}\)2. \(\frac{{{b^2}\left( {2{b^2} + {a^2}} \right)}}{{{a^2}\left( {{a^2} + {b^2}} \right)}}\)3. \(\frac{{{a^2}\left( {2{b^2} - {a^2}} \right)}}{{{b^2}\left( {{a^2} + {b^2}} \right)}}\)4. \(\frac{{{a^2}\left( {2{b^2} - {a^2}} \right)}}{{{a^2}\left( {{a^2} - {b^2}} \right)}}\)
Answer» Correct Answer - Option 3 : \(\frac{{{a^2}\left( {2{b^2} - {a^2}} \right)}}{{{b^2}\left( {{a^2} + {b^2}} \right)}}\) Given- secθ = a/b Concept Used- secθ = H/B, tanθ = P/B, sinθ = P/H and H² = P² + B² [where H = hypotenuse, B = base and P = perpendicular] Calculation- According to Question - H/B = a/b P = √(a² - b²), H = a and B = b tan²θ = (a2 - b2)/b² sin²θ = (a2 - b2)/a2 (1 - tan²θ)/(2 - sin²θ) ⇒ {(2b² - a²)/b²}/{(a² + b²)/a²} \(\Rightarrow \frac{{{a^2}\left( {2{b^2} - {a^2}} \right)}}{{{b^2}\left( {{a^2} + {b^2}} \right)}}\) ∴\(\frac{{1 - {{\tan }^2}\theta }}{{2 - {{\sin }^2}\theta }}=\frac{{{a^2}\left( {2{b^2} - {a^2}} \right)}}{{{b^2}\left( {{a^2} + {b^2}} \right)}}\)

If \(sec\theta = \frac{a}{b},b \ne 0,then\frac{{1 - {{\tan }^2}\theta }}{{2 - {{\sin }^2}\theta }}\)=?1. \(\frac{{{b^2}\left( {2{b^2} - {a^2}} \right)}}{{{a^2}\left( {{a^2} + {b^2}} \right)}}\)2. \(\frac{{{b^2}\left( {2{b^2} + {a^2}} \right)}}{{{a^2}\left( {{a^2} + {b^2}} \right)}}\)3. \(\frac{{{a^2}\left( {2{b^2} - {a^2}} \right)}}{{{b^2}\left( {{a^2} + {b^2}} \right)}}\)4. \(\frac{{{a^2}\left( {2{b^2} - {a^2}} \right)}}{{{a^2}\left( {{a^2} - {b^2}} \right)}}\)

Answer» Correct Answer - Option 3 : \(\frac{{{a^2}\left( {2{b^2} - {a^2}} \right)}}{{{b^2}\left( {{a^2} + {b^2}} \right)}}\)

Given-

secθ = a/b

Concept Used-

secθ = H/B, tanθ = P/B, sinθ = P/H and H² = P² + B² [where H = hypotenuse, B = base and P = perpendicular]

Calculation-

According to Question -

H/B = a/b

P = √(a² - b²), H = a and B = b

tan²θ = (a2 - b2)/b²

sin²θ = (a2 - b2)/a2

(1 - tan²θ)/(2 - sin²θ)

⇒ {(2b² - a²)/b²}/{(a² + b²)/a²}

\(\Rightarrow \frac{{{a^2}\left( {2{b^2} - {a^2}} \right)}}{{{b^2}\left( {{a^2} + {b^2}} \right)}}\)

∴\(\frac{{1 - {{\tan }^2}\theta }}{{2 - {{\sin }^2}\theta }}=\frac{{{a^2}\left( {2{b^2} - {a^2}} \right)}}{{{b^2}\left( {{a^2} + {b^2}} \right)}}\)