Question

Q5.If cot Ø + cos Ø = x and cot Ø - cos Ø =y then x²-y² = ​

Answer 1

Answer:4cospcotp

Step-by-step explanation:let phi=p

Given, x=cotp+cosp

Y=cotp-cosp

X^2=cot^2p+ cos^2p+ 2cospcotp

Y^.2 =cos^2p+ cot^2p+ 2cospcotp

X^2-Y^2. =4cospcotp

Answer 2

$\sf\red{\underline{\underline{Answer:}}}$

$\sf{The \ value \ of \ x^{2}-y^{2} \ is \ 4cot\theta.cos\theta}$

$\sf\orange{Given:}$

$\sf{\implies{cot\theta+cos\theta=x}}$

$\sf{\implies{cot\theta-cos\theta=y}}$

$\sf\pink{To \ find:}$

$\sf{x^{2}-y^{2}}$

$\sf\green{\underline{\underline{Solution:}}}$

$\sf{According \ to \ the \ identity}$

$\sf{a^{2}-b^{2}=(a+b)(a-b)}$

$\sf{x^{2}-y^{2}=(x+y)(x-y)}$

$\sf{x^{2}-y^{2}}$

$\sf{=[(cot\theta+cos\theta)+(cot\theta-cos\theta)][(cot\theta+cos\theta)-(cot\theta-cos\theta)]}$

$\sf{x^{2}-y^{2}=(2cot\theta)(2cos\theta)}$

$\sf{\therefore{x^{2}-y^{2}=4cot\theta.cos\theta}}$

$\sf\purple{\tt{\therefore{The \ value \ of \ x^{2}-y^{2} \ is \ 4cot\theta.cos\theta}}}$