Question

Solve: √1+cos x÷√1-cos x + √1-cos x ÷ √1+cos x

Answer 1

Answer:

LHS = 1/sin x(1/cos x-1) - cos x/sinx (1-cos x)

=1-cos x/sin xcos x -( 1-cosx) cosx/sinx

=(1-cosx/sinx)[1/cosx - cosx]

=(1-cosx/sinx)sin*2x/cosx

=(1-cosx)sinx/cosx

=sinx/cosx -sinx

=tanx-sinx=RHS

HENCE LHS=RHS

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Answer 2

$\Large\underline\mathfrak{Solve}$ :---

$\green{\boxed{\bf \: \sqrt{\frac{1+cosx}{1-cosx}} + \sqrt{\frac{1 - cosx}{1 + cosx}}}}$

$\Large\bold\star\underline{\underline\textbf{Formula\:used}}$

• (a+b)(a-b) = a² - b²
• 1 - cos²x = sin²x
• √(a)² = a

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$\bigstar\underline{\red{\mathbb{LET'S \: \color{lime}{SOLVE} \: \color{fuchsia}{THE}\: \color{aqua}{PROBLEM}}}}\bigstar \:$

$\Large\bold\star\underline{\underline\textbf{Part(1)}}$

$\sf \: \sqrt{\frac{1+cosx}{1-cosx}} \: = \\ \\ \orange{\sf \: multiply \: and \: divide \: by \: ( 1+ cosx)we \: get} \\ \\\red\leadsto\rm\sqrt{\frac{1+cosx}{1-cosx} \times \frac{1 + cosx}{1 + cosx}} \\ \\ \\\red\leadsto\rm \: \sqrt{ \frac{(1 + cosx)^{2} }{1 - cos^{2}x } } \\ \\ \\\red\leadsto \pink{\boxed{\bf \: \dfrac{1 + cosx}{sinx}}}$

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$\Large\bold\star\underline{\underline\textbf{Part(2)}}$

$\sf \: \sqrt{\frac{1 - cosx}{1 + cosx}} \: = \\ \\ \blue{\sf \: multiply \: and \: divide \: by \: ( 1 - cosx) \: we \: get} \\ \\\red\leadsto\rm\sqrt{\frac{1 - cosx}{1 + cosx} \times \frac{1 - cosx}{1 - cosx}} \\ \\ \\\red\leadsto\rm \: \sqrt{ \frac{(1 - cosx)^{2} }{1 - cos^{2}x } } \\ \\ \\\red\leadsto \purple{\boxed{\bf \: \dfrac{1 - cosx}{sinx}}} \:$

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$\red{\underline\textbf{Adding Part(1) and Part(2) Now, we get,}}$

$\red\leadsto \sf \dfrac{1 + cosx}{sinx} + \dfrac{1 - cosx}{sinx} \\ \\ \red\leadsto \rm \: \dfrac{1 + \cancel{cosx} + 1 - \cancel{cosx}}{sinx} \\ \\ \red\leadsto \: \bold{\boxed{\large{\boxed{\orange{\small{\boxed{\large{\red{\bold{\: \frac{2}{sinx} }}}}}}}}}}$

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