Prove the following trigonometric identities. If x a + y b =1 and y a - y b =1, prove that x^2 a^2 + y^2 b^2 =2.

Given that, x a + y b =1 .....(i) x a - y b =1 .....(ii) We have to prove x^2 a^2 + y^2 b^2 =2 We know that, ^2 + ^2 =1 Squaring and then adding the above two equations, we have L.H.S= ( x a + y b )^2+ ( x a - y b )^2=1+1 ( x^2 a^2 ^2 +2 xy ab + y^2 b^2 ^2 ) = ( x^2 a^2 ^2 -2 xy ab + y^2 b^2 ^2 )=2 x^2 a^2 ( ^2 + ^2 )+ y^2 b^2 ( ^2 + ^2 )=2 x^2 a^2 + y^2 b^2 =2=R.H.S L.H.S=R.H.S

Prove the following trigonometric identities. If x a + y b =1 and…

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Question

Prove the following trigonometric identities.
If $\frac{\text{x}}{\text{a}}\cos\theta+\frac{\text{y}}{\text{b}}\sin\theta=1\text{ and }\frac{\text{y}}{\text{a}}\sin\theta-\frac{\text{y}}{\text{b}}\cos\theta=1,$ prove that $\frac{\text{x}^2}{\text{a}^2}+\frac{\text{y}^2}{\text{b}^2}=2.$

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Answer

Given that,
$\frac{\text{x}}{\text{a}}\cos\theta+\frac{\text{y}}{\text{b}}\sin\theta=1\ .....\text{(i)}$
$\frac{\text{x}}{\text{a}}\sin\theta-\frac{\text{y}}{\text{b}}\cos\theta=1\ .....\text{(ii)}$
We have to prove $\frac{\text{x}^2}{\text{a}^2}+\frac{\text{y}^2}{\text{b}^2}=2$
We know that, $\sin^2\theta+\cos^2\theta=1$
Squaring and then adding the above two equations, we have
$\text{L.H.S}=\Big(\frac{\text{x}}{\text{a}}\cos\theta+\frac{\text{y}}{\text{b}}\sin\theta\Big)^2+\Big(\frac{\text{x}}{\text{a}}\sin\theta-\frac{\text{y}}{\text{b}}\cos\theta\Big)^2=1+1$
$\Rightarrow \Big(\frac{\text{x}^2}{\text{a}^2}\cos^2\theta+2\frac{\text{xy}}{\text{ab}}\sin\theta\cos\theta+\frac{\text{y}^2}{\text{b}^2}\sin^2\theta\Big)$
$=\Big(\frac{\text{x}^2}{\text{a}^2}\sin^2\theta-2\frac{\text{xy}}{\text{ab}}\sin\theta\cos\theta+\frac{\text{y}^2}{\text{b}^2}\cos^2\theta\Big)=2$
$\Rightarrow\ \frac{\text{x}^2}{\text{a}^2}(\cos^2\theta+\sin^2\theta)+\frac{\text{y}^2}{\text{b}^2}(\sin^2\theta+\cos^2\theta)=2$
$\Rightarrow \frac{\text{x}^2}{\text{a}^2}+\frac{\text{y}^2}{\text{b}^2}=2=\text{R.H.S}$
$\therefore\ \text{L.H.S}=\text{R.H.S}$