Question

$\huge{\underline{\underline{\boxed{\sf{\purple{ǫᴜᴇsᴛɪᴏɴ}}}}}}$

$If \: sinθ \: = \frac{a}{b} \: \\ then \: prove \: that \: (secθ + tanθ) = \sqrt{ \frac{b + a}{b - a} }$
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Answer 1

Solution:-

Given

$\rm \implies \: \sin \theta = \dfrac{a}{b}$

We have to proof

$\implies \sec \theta \: + \tan \theta = \sqrt{ \rm\dfrac{b + a}{b - a} }$

Now take

$\rm \implies \: \sin \theta = \dfrac{a}{b} = \dfrac{p}{h}$

Using Pythagoras theorem we will find Base(x)

$\rm \implies \: {h}^{2} = {x}^{2} + {p}^{2}$

$\rm \implies \: {b}^{2} = {x}^{2} + {a}^{2}$

$\rm \implies \: {x}^{2} = {b}^{2} - {a}^{2}$

$\rm \implies \: x = \sqrt{ {b}^{2} - {a}^{2} }$

Now we know that

$\rm \implies \: \sec \theta = \dfrac{h}{x} = \dfrac{b}{ \sqrt{ {b}^{2} - {a}^{2} } }$

$\implies \rm \tan \theta = \dfrac{p}{x} = \dfrac{a}{ \sqrt{ {b}^{2} - {a}^{2} } }$

Now take

$\rm \implies \sec \theta \: + \tan \theta$

$\rm \implies \: \dfrac{b}{ \sqrt{ {b}^{2} - {a}^{2} } } + \dfrac{a}{ \sqrt{ {b}^{2} - {a}^{2} } }$

$\rm \implies \: \dfrac{b + a}{ \sqrt{ {b}^{2} - {a}^{2} } }$

$\rm \implies \: \dfrac{b + a}{ \sqrt{ {b - a} } \sqrt{b + a} }$

$\rm \implies \: \dfrac{b + a}{ \sqrt{ {b - a} } \sqrt{b + a} } \times \dfrac{ \sqrt{b + a} }{ \sqrt{b + a} }$

$\rm \implies \: \dfrac{b + a \sqrt{b + a} }{b + a \sqrt{b - a} }$

$\rm \implies \: \dfrac{ \cancel{b + a} \sqrt{b + a} }{ \cancel{b + a} \sqrt{b - a} }$

$\rm \implies \: \dfrac{ \sqrt{b + a} }{ \sqrt{b - a} } = \sqrt{ \dfrac{b + a}{b - a} }$

Hence proved

Answer 2

Answer:

Given :-

➙ sin$\theta$ = $\dfrac{a}{b}$

Prove That :-

➙ (sec$\theta$ + tan$\theta$ = $\sqrt{\dfrac{b + a}{b - a}}$

Solution :-

➤ L.H.S = (sec$\theta$ + tan$\theta$)

↪ $\dfrac{1}{cos\theta}$ + $\dfrac{sin\theta}{cos\theta}$

↪ $\dfrac{1 + sin\theta}{cos\theta}$

↪ $\dfrac{1 + sin\theta}{\sqrt{1 - {sin}^{2}\theta}}$

↪ $\dfrac{(1 + \dfrac{a}{b})}{\sqrt{1 + (\dfrac{a}{b})^{2}}}$

↪ $\dfrac{(\dfrac{1}{1} + \dfrac{a}{b})}{\sqrt{\dfrac{1}{1} - \dfrac{a²}{b²}}}$

↪ $\dfrac{(\dfrac{b + a}{b})}{\sqrt{\dfrac{b² - a²}{b²}}}$

↪ $\dfrac{(b + a)}{{\sqrt{b + a}}{\sqrt{b - a}}}$

↪ $\dfrac{\sqrt{(b + a)}}{\sqrt{(b - a)}}$

↪ $\sqrt{\dfrac{b + a}{b - a}}$

➤ R.H.S

$\leadsto$ $\large\green{\underline{{\boxed{\textbf{PROVED}}}}}$