Question

[secA- tanA]²= 1-SinA / 1+sinA
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Answer 1

Answer:

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Answer 2

$\bf\red {\underline {We\:Have\:To\:Prove:-}}$

• $\tt {( secA - tanA)^{2} = \frac{1 - sinA}{1 +sinA}}$

$\bf\pink{\underline {Solution:-}}$

$\tt {Left \: Side}$

$\tt { = ( secA - tanA)^{2} }$

$\tt { = {sec}^{2}A - 2sec AtanA + {tan}^{2} A}$

$\tt {= \frac{1}{ {cos}^{2} A} - 2 \times \frac{1}{cosA} \times \frac{sinA}{cosA} + \frac{ {sin}^{2} A}{ {cos}^{2}A }}$

$\tt{ = \frac{1}{ {cos}^{2}A } - \frac{2sinA}{ {cos}^{2} A} + \frac{ {sin}^{2}A }{ {cos}^{2} A}}$

$\tt {= \frac{1 - 2sinA + {sin}^{2}A }{ {cos}^{2}A } }$

$\tt {= \frac{1 - sinA - sinA + {sin}^{2}A }{1 - {sin}^{2}A} }$

$\tt {= \frac{(1 - sinA) - sinA(1 - sinA)}{(1 + sinA)(1 - sinA)}}$

$\tt{= \frac{(1 - sinA)(1 - sinA)}{(1 + sinA)(1 - sinA)}}$

$\tt{= \frac{1 - sinA}{1 +sinA } }$

$\tt{= Right \: Side}$

$\bf\green{[Hence\:Proved]}$

Important Formulas :-

• $\bf\blue{{sin}^{2} θ + {cos}^{2} θ = 1}$
• $\bf\blue{{sec}^{2} θ - {tan}^{2} θ = 1}$
• $\bf\blue{{cosec}^{2} θ - {cot}^{2} θ = 1}$