Question

Prove the given trigonometry identity :-
√(1 + sin A/1 – sin A) = sec A + tan A
No Copy or Spam

Answer 1

$\tt To\:prove \: : \sqrt{\frac{1+sinA}{1-sinA}} = secA + tanA\\ \\ \tt L.H.S \\ \\ \tt \implies \sqrt{\frac{1+sinA}{1-sinA}} \\ \\ \tt Dividing\:and\: multiplying\:by (1+sinA) \\ \\ \tt \implies \sqrt{\frac{(1+sinA)(1+sinA)}{(1+sinA)(1-sinA)}} \\ \\ \tt \implies \sqrt{\frac{{(1+sinA)}^{2}}{1-{sin}^{2}A}} \\ \\ \tt \implies \sqrt{\frac{{(1+sinA)}^{2}}{{cos}^{2}A}} \\ \\ \tt \implies \frac{1+sinA}{cosA} \\ \\ \tt \implies \frac{1}{cosA} + \frac{sinA}{cosA} \\ \\ \tt \implies secA + tanA \\ \\ \tt secA+ tanA \\ \\ \tt \implies R.H.S \\ \\ \underline{\red{\tt Hence\:proved}} \large{\purple{\mathfrak{Formulas\:and\: identities}}} \\ \\ \orange{\mathbb{ IDENTITIES }} \\ \\ \tt \circ \:\: {sin}^{2}\theta + {cos}^{2}\theta = 1 \\ \tt \circ \:\: 1+{tan}^{2}\theta = {sec}^{2}\theta \\ \tt \circ \:\: 1 + {cot}^{2}\theta = {cosec}^{2}\theta \\ \\ \orange{\mathbb{ DOUBLE\:ANGLE\:FORMULAS}} \\ \\ \tt \circ \:\: sin2\theta = 2sin\theta cos\theta \\ \tt \circ \:\: cos2\theta = {cos}^{2}\theta - {sin}^{2}\theta \\ \tt \circ \:\: tan2\theta = \frac{2tan\theta}{1-{tan}^{2}\theta} \\ \\ \orange{\mathbb{TRIPLE\:ANGLE\:FORMULAS}} \\ \\ \tt \circ \:\: sin3\theta = 3sin\theta - 4{sin}^{3}\theta \\ \tt \circ \:\: cos3\theta = 4{cos}^{3}\theta - 3cos\theta \\ \tt \circ \:\: tan3\theta = \frac{3tan\theta - {tan}^{3}\theta}{1-3{tan}^{2}\theta} \\ \\ \orange{\mathbb{GENERAL\: TRANSFORMATIONS}} \\ \\ \tt \circ \:\: cosec\theta = \frac{1}{sin\theta} \\ \tt \circ \:\: sec\theta = \frac{1}{cos\theta} \\ \tt \circ \:\: cot\theta = \frac{1}{tan\theta}$

Answer 2

Question:-

Prove the given trigonometry identity :-

√(1 + sin A/1 – sin A) = sec A + tan A

Required Answer:-

Given:-

$\\\sf{Prove \: that: { \sqrt{ \dfrac{1 + sinA}{1 - sinA }} = secA + tanA}}$

Identity used:-

$\\ \sf{\rightarrow{1 - {sin}^{2}\theta = {cos}^{2} \theta}}$

$\sf{\rightarrow{(a + b)(a - b) = {a}^{2} - {b}^{2}}}$

Solution:-

L.H.S.:-

$\\\sf{\bold{ \sqrt{ \dfrac{1 + sinA}{1 - sinA }}}}$

Rationalising the denominator we get:-

$\\ \sf{\implies{ \sqrt{ \dfrac{(1 + sinA)(1 + sinA)}{((1 - sinA)(1 + sinA)}}} }$

$\\ \sf{\implies{ \sqrt{ \frac{ {(1 + sinA)}^{2} }{ {1}^{2} - {sin}^{2}A }}}}$

$\\ \sf{\implies{ \sqrt{ \dfrac{ {(1 + sinA)}^{2} }{ {cos}^{2}A }}}} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \boxed{\rm{\because{ {1 - {sin}^{2}\theta = {cos}^{2} \theta}}}}$

$\\ \sf{\implies{ \dfrac{1 + sinA}{cosA}}}$

$\\ \sf{\implies{ \dfrac{1}{cosA} + \dfrac{sinA}{cosA}}}$

Now, applying 1/cosA=secA and sinA/cosA=tanA we get:-

$\\ \sf{\implies{\red{secA + tanA}}}$

=> R.H.S.

We observe that L.H.S. = R.H.S.

Hence, proved!

Some Important Identities:-

$\\ \sf{\rightarrow{ {sin}^{2} \theta + {cos}^{2} \theta = 1}} \\ \\ \sf{\rightarrow{ {sec}^{2}\theta = 1 + {tan}^{2} \theta}} \\ \\ \sf{\rightarrow{ {cosec}^{2} \theta = 1 + {cot}^{2}\theta}}$