Math, asked by arnab2261, 10 months ago

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Answered by Anonymous

Question:

$\sqrt{\frac{(1+sin2\theta)}{1-cos^2\theta}}$

Here θ ∈ [ 0 , π/4 ]

OPTION ( a ) cosec²θ

OPTION ( b ) 1

OPTION ( c ) 1 + cotθ

OPTION ( d ) 1 + tanθ

Answer:

$\huge\boxed{\boxed{\red{OPTION\:C}}}$

Step-by-step explanation:

The given expression is :-

$\sqrt{\frac{(1+sin2\theta)}{1-cos^2\theta}}$

$sin2\theta\textbf{ can be written as }2sin\theta cos\theta\\\\1+sin2\theta\textbf{ can hence be written as }1+2sin\theta cos\theta\\\\\implies sin^2\theta+cos^2\theta+2sin\theta cos\theta\\\\\implies sin^2\theta+sin\theta cos\theta +sin\theta cos\theta+ cos^2\theta\\\\\implies sin\theta(sin\theta+cos\theta)+cos\theta(sin\theta+cos\theta)\\\\\implies (sin\theta+cos\theta)(sin\theta+cos\theta)$

$\sqrt{\frac{1+sin2\theta}{1-cos^2\theta}}\\\\\implies \sqrt{\frac{(sin\theta+cos\theta)(sin\theta+cos\theta)}{1-cos^2\theta}}\\\\\textbf{Note that $1-cos^2\theta=sin^2\theta$}\\\\\implies \sqrt{\frac{(sin\theta+cos\theta)^2}{sin^2\theta}}\\\\\implies \frac{sin\theta+cos\theta}{sin\theta}\\\\\implies \frac{sin\theta}{sin\theta}+\frac{cos\theta}{sin\theta}\\\\\implies 1+cot\theta\\\\\bf{I\:used\:\frac{cos\theta}{sin\theta}=cot\theta}$

So the correct answer would be OPTION C :)

The formulas that were used in the above problem :-

$\boxed{\bf{sin2x=2sinxcosx}}\\\\\boxed{\bf{sin^2x+cos^2x=1}}\\\\\textbf{All-sin-tan-cos-rule is used here.}\\\\\textsf{All trigonometric ratios are positive in the first quadrant.}$

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Answered by Anonymous

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