Question

If cot theta =13/7 then find the value of (2+2 sin theta) ( 1- sin theta)/ (1+cos theta) (2-2 cos theta)​

Answer 1

ANSWER:

Given:

• cotΘ = 13/7

To Find:

• Value of [(2 + 2sinΘ)(1 - sinΘ)]/[(1 + cosΘ)(2 - 2cosΘ)]

Solution:

$:\longrightarrow\dfrac{(2+2\sin\theta)(1-\sin\theta)}{(1+\cos\theta)(2-2\cos\theta)}\\\\\text{Taking 2 common,}\\\\:\implies\dfrac{2\!\!\!/(1+\sin\theta)(1-\sin\theta)}{2\!\!\!/(1+\cos\theta)(1-\cos\theta)}\\\\\text{We know that, (a+b)(a-b)=a^2-b^2 . So,}\\\\:\implies\dfrac{1^2-\sin^2\theta}{1^2-\cos^2\theta}\\\\\text{Also, \sin^2\theta+\cos^2\theta=1 . So, by rearranging it,}\\\\:\longrightarrow1-\sin^2\theta=\cos^2\theta\:\:\:\:\:\&\:\:\:\:\:1-\cos^2\theta=\sin^2\theta.\:\:\:\:\text{So,}$

$:\implies\dfrac{1-\sin^2\theta}{1-\cos^2\theta}\\\\:\implies\dfrac{\cos^2\theta}{\sin^2\theta}\\\\:\implies\left(\dfrac{\cos\theta}{\sin\theta}\right)^{2}\\\\\text{We know that, \dfrac{\cos\theta}{\sin\theta} = \cot\theta . So,}\\\\:\implies\left(\cot\theta\right)^2\\\\\text{We are given that, \cot\theta=\dfrac{13}{7} . Hence,}\\\\:\implies\left(\dfrac{13}{7}\right)^2\\\\:\implies\dfrac{13^2}{7^2}\\\\\bf{:\implies\dfrac{169}{49}}$

Formulae Used:

• (a + b)(a - b) = a² - b²
• sin²Θ + cos²Θ = 1
• cosΘ/sinΘ = cotΘ