Question

If cot Φ = 3/4,prove that √secΦ-cosecΦ/secΦ+cosecΦ =1/√7.

Answer 1

QUESTION :

If $\sf{cot\theta=\frac{3}{4}}$

Prove that,

$\sf{\sqrt{\frac{sec\theta-cosec\theta}{sec\theta+cosec\theta}}=\frac{1}{\sqrt{7}}}$

SOLUTION :

Given:-

$\sf{cot\theta=\frac{3}{4}}$

We know,

$\sf{cot\theta=\frac{Base}{Height}=\frac{3}{4}}$

According to "Pythagoras Theorem " :

Height ² + Base² = Hypotenuse ²

→4² + 3² = Hypotenuse ²

→16+9 = Hypotenuse ²

→ 25 = Hypotenuse ²

→ 5 = Hypotenuse

So,

★$\sf{sec\theta=\frac{Hypotenuse}{Base}=\frac{5}{3}}$

★$\sf{cosec\theta=\frac{Hypotenuse}{Height}=\frac{5}{4}}$

Taking L.H.S,

$\sf{\sqrt{\frac{sec\theta-cosec\theta}{sec\theta+cosec\theta}}}$

$\implies\sf{\sqrt{\frac{\frac{5}{3}-{\frac{5}{4}}}{\frac{5}{3}+{\frac{5}{4}}}}}$

$\implies\sf{\sqrt{\frac{\frac{5}{12}}{\frac{35}{12}}}}$

$\implies\sf{\sqrt{\frac{5}{12}\times{\frac{12}{35}}}}$

$\implies\sf{\sqrt{\frac{1}{7}}}$

$\implies\sf{\frac{1}{\sqrt{7}}}$

L.H.S = R.H.S ( Proved)

Answer 2

Given:-

$\sf{cot\theta=\frac{3}{4}}$

We know,

$\sf{cot\theta=\frac{Base}{Height}=\frac{3}{4}}$

According to "Pythagoras Theorem " :

Height ² + Base² = Hypotenuse ²

→4² + 3² = Hypotenuse ²

→16+9 = Hypotenuse ²

→ 25 = Hypotenuse ²

→ 5 = Hypotenuse

So,

$\sf{sec\theta=\frac{Hypotenuse}{Base}=\frac{5}{3}}$

$\sf{cosec\theta=\frac{Hypotenuse}{Height}=[tex]\frac{5}{4}}$

Taking L.H.S,

$\sf{\sqrt{\frac{sec\theta-cosec\theta}{sec\theta+cosec\theta}}}$

$\implies\sf{\sqrt{\frac{\frac{5}{3}-{\frac{5}{4}}}{\frac{5}{3}+{\frac{5}{4}}}}}$

$\implies\sf{\sqrt{\frac{\frac{5}{12}}{\frac{35}{12}}}}$

$\implies\sf{\sqrt{\frac{5}{12}\times{\frac{12}{35}}}}$

$\implies\sf{\sqrt{\frac{1}{7}}}$

$\implies\sf{\frac{1}{\sqrt{7}}}$

L.H.S = R.H.S ( Proved)