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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. $\frac { 1 + \sec A } { \sec A } = \frac { \sin ^ { 2 } A } { 1 - \cos A }$
[Hint: Simplify LHS and RHS separately]
✓

Answer

$\mathrm { LHS } = \frac { 1 + \sec A } { \sec A } = \frac { 1 + \frac { 1 } { \cos A } } { \frac { 1 } { \cos A } }$
$= \frac { \frac { \cos A + 1 } { \cos A } } { \frac { 1 } { \cos A } } = \cos A + 1 = 1 + \cos A$
$= \frac { ( 1 + \cos A ) ( 1 - \cos A ) } { 1 - \cos A } = \frac { 1 - \cos ^ { 2 } A } { 1 - \cos A }$
$= \frac { \sin ^ { 2 } A } { 1 - \cos A } \cdot \because \sin ^ { 2 } A + \cos ^ { 2 } A = 1$
= RHS

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