Question

If $cot\Theta =\frac{3}{4}$, prove that $\frac{\sqrt{sec\Theta}-cosec\Theta }{sec\Theta+cosec\Theta}=\frac{1}{\sqrt{7} }$

Answer 1

SOLUTION :  

Given: cot θ =¾

cot θ = ¾ = Base / perpendicular = AB / BC

Base  = 3 and Hypotenuse = 4

In right angled ΔABC, by using Pythagoras theorem

AC² = AB² + BC²

AC² = 3² + 4²

AC² = 9 + 16

AC² = 25

AC = √25

AC = 5

Hypotenuse (AC) = 5

Now, cosec θ = Hypotenuse/ Perpendicular  

cosec  θ = 5/4

sec θ = Hypotenuse / base

sec θ = 5/3

CALCULATION of prove that is in the ATTACHMENT.

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Answer 2

cotΘ = $\dfrac{3}{4}$

We know that cotA is the ratio of base of the triangle to its height. Therefore, cotA = $\dfrac{Base}{Height}$

Now,

⇒$\dfrac{Base}{Height} = \dfrac{3}{4}$

⇒ $\dfrac{\dfrac{Base}{Hypotenuse}}{\dfrac{Height}{Hypotenuse}} = \dfrac{3}{4}$

Note that the ratio of base to hypotenuse with reference to any particular angle is cosA and the ratio of height to hypotenuse with reference to any particular angle is sinA, where A is that angle.

⇒ $\dfrac{cos \Theta}{sin \Theta}= \dfrac{3}{4}$

We know that sinA is the reciprocal of cosecA and reciprocal of secA is reciprocal of cosA.

∴ sinA = $\dfrac{1}{cosecA}\:\:\:\and\:\:\:\:\:cosA =\dfrac{1}{secA}$

⇒ $\dfrac{cosec\Theta}{sec\Theta} = \dfrac{3}{4}$

⇒ $\dfrac{4}{3} = \sqrt{ \dfrac{sec \Theta}{cosec \Theta}$

By Componendo & Dividendo,

⇒ $\dfrac{4 + 3}{ 4 - 3} =\sqrt{ \dfrac{sec\Theta + cosec\Theta}{secA - cosecA}}$

⇒ $\dfrac{7}{1} = \sqrt{\dfrac{secA + cosecA}{secA - cosecA}}$

⇒ $\dfrac{1}{7} = [tex]\sqrt{\dfrac{secA - cosecA}{secA + cosecA}}$

Square root on both sides,

⇒ $\dfrac{1}{\sqrt{7}} = <strong>\sqrt{\dfrac{secA - cosecA}{secA + cosecA}}$

Hence, If  $cot\Theta =\frac{3}{4},[\tex] prove that \frac{\sqrt{sec\Theta}-cosec\Theta }{sec\Theta+cosec\Theta}=\frac{1}{\sqrt{7} }$