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Introduction of Trigonometry — Maths STD 10 — Question

CBSE BoardEnglish MediumSTD 10MathsIntroduction of Trigonometry1 Mark
Question
Prove the given identity, where the angles involved are acute angles for which the expressions are defined. $\sqrt { \frac { 1 + \sin A } { 1 - \sin A } }$ = sec A + tan A
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Answer

L.H.S. $\sqrt { \frac { 1 + \sin A } { 1 - \sin A } }$
$= \sqrt { \frac { 1 + \sin A } { 1 - \sin A } } \times \sqrt { \frac { 1 + \sin A } { 1 + \sin A } }$
$=\sqrt { \frac { ( 1 + \sin A ) ^ { 2 } } { 1 - \sin ^ { 2 } A } } \left[ \because ( a + b ) ( a - b ) = a ^ { 2 } - b ^ { 2 } \right]$
$=\sqrt { \frac { ( 1 + \sin A ) ^ { 2 } } { \cos ^ { 2 } A } } \left[ \because 1 - \sin ^ { 2 } \theta = \cos ^ { 2 } \theta \right]$
$=\frac { 1 + \sin A } { \cos A } = \frac { 1 } { \cos A } + \frac { \sin A } { \cos A } = \sec A + \tan A = R \cdot H . S .$

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