यदि \(\sin θ + \cos θ = \frac{{\sqrt 3 - 1}}{{2\sqrt 2 }}\) है, तो tan θ + cot θ का मान क्या है?

Question

Question

This question was previously asked in

SSC CGL 2022 Tier-I Official Paper (Held On : 05 Dec 2022 Shift 1)

Answer 1

दिया गया है:

\(\sin θ + \cos θ = \frac{{\sqrt 3 - 1}}{{2\sqrt 2 }}\)

प्रयुक्त अवधारणा:

sin²θ + cos²θ = 1

गणना:

\(\sin θ + \cos θ = \frac{{\sqrt 3 - 1}}{{2\sqrt 2 }}\)

⇒ \((\sin θ + \cos θ)^2 = \left(\frac{{\sqrt 3 - 1}}{{2\sqrt 2 }}\right)^2\)

⇒ \(\sin ^2θ + \cos ^2θ +2sinθ .cosθ = \left(\frac{{3 + 1-2\sqrt3.1}}{{8 }}\right)\)

⇒ \(1+2sinθ .cosθ = \frac{{4-2\sqrt3}}{{8 }}\)

⇒ \(2sinθ .cosθ = \frac{{4-2\sqrt3}}{{8 }}-1\)

⇒ \(2sinθ .cosθ = \frac{{4-2\sqrt3-8}}{{8 }}\)

⇒ \(sinθ .cosθ = \frac{{-(\sqrt3+2)}}{{8 }}\)

अब,

tan θ + cot θ = \(\rm \frac{sinθ}{cosθ}+\frac{cosθ}{sinθ}\)

⇒ \(\rm \frac{sin^2θ+cos^2θ}{sinθ.cosθ}\)

⇒ \(\rm \frac{1}{sinθ.cosθ}\)

⇒ \(\rm \frac{1}{\frac{{-(\sqrt3+2)}}{{8 }}}\)

⇒ \(\rm \frac{-8}{{{(\sqrt3+2)}}{}}\)

⇒ \(\rm \frac{-8(\sqrt3-2)}{{{(\sqrt3+2)(\sqrt3-2)}}{}}\)

⇒ \(\rm \frac{-8(\sqrt3-2)}{{{3-4}}{}}\)

⇒ \(\rm \frac{-8(\sqrt3-2)}{{{-1}}{}}\)

⇒ \(8\left( {\sqrt3 - 2} \right)\)

∴ अभीष्ट उत्तर \(8\left( {\sqrt3 - 2} \right)\) है।