(sec 2A + 1){sec ^2}A =

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3.Trigonometrical Ratios, Functions and Identities

easy

$(\sec 2A + 1){\sec ^2}A = $

A

$\sec A$

B

$2\sec A$

C

$\sec 2A$

D

$2\sec 2A$

Solution

(d) $(\sec 2A + 1){\sec ^2}A $

$= \left( {\frac{{1 + {{\tan }^2}A}}{{1 – {{\tan }^2}A}} + 1} \right)\,(1 + {\tan ^2}A)$

$ = \frac{{2\,(1 + {{\tan }^2}A)}}{{1 – {{\tan }^2}A}}$

$= 2\sec 2A.$

Standard 11

Mathematics

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