यदि tan² α = 3 + Q ² , sec α + tan³ α cosec α = ?

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Question

This question was previously asked in

SSC CGL 2022 Tier-I Official Paper (Held On : 07 Dec 2022 Shift 1)

Answer 1

दिया गया:

tan2 \(\alpha\) = 3 + Q2

प्रयुक्त सूत्र:

Sec2 \(\alpha\) - tan2 \(\alpha\) = 1

गणना:

⇒ tan2 \(\alpha\) = 3 + Q2

⇒ Sec2 \(\alpha\) - 1 = 3 + Q2

⇒ Sec2 \(\alpha\) = 4 + Q2

⇒ Sec \(\alpha\) = \(\sqrt{4+Q^2}\)

दिया गया समीकरण है:

⇒ Sec \(\alpha\) + tan3 \(\alpha\) cosec \(\alpha\)

\(\frac{1}{cos \alpha} + \frac{ sin^3 \alpha}{cos^3 \alpha}\times\frac{1}{sin \alpha}\)

\(\frac{1}{cos \alpha} + \frac{ sin^2 \alpha}{cos^3 \alpha} \)

\(\frac{cos^2\alpha +sin^2 \alpha}{cos^3 \alpha} \)

\(\frac{1}{cos^3 \alpha} \)

⇒ Sec3 \(\alpha\)

(Sec \(\alpha\))3

\((\sqrt{4+Q^2})^3\)

\((4+Q^2 )^{\frac{3}{2}}\)

⇒ अतः, विकल्प 4 सही है।

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