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

x=a x a = .....i y=b y b = .....(ii) z c = .....(iii) We have to prove that x^2 a^2 + y^2 b^2 - z^2 c^2 =1 Squaring the above equations and then subtracting the third from the sum of the first two, we have ( x a )^2+ ( y b )^2- ( z c )^2=( )^2+( )-( )^2 x^2 a^2 + y^2 b^2 + z^2 c^2 = ^2 ^2 + ^2 ^2 - ^2 x^2 a^2 + y^2 b^2 + z^2 c^2 =( ^2 ^2 + ^2 ^2 )- ^2 x^2 a^2 + y^2 b^2 + z^2 c^2 = ^2 ( ^2 + ^2 )- ^2 x^2 a^2 + y^2 b^2 + z^2 c^2 = ^2 (1)- ^2 x^2 a^2 + y^2 b^2 + z^2 c^2 = ^2 - ^2 x^2 a^2 + y^2 b^2 + z^2 c^2 =1 Hence proved.

Prove the following trigonometric identities. If x=a ,y=b and z=c…

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Question

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

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Answer

$\text{x}=\text{a}\sec\theta\cos\phi$
$\Rightarrow\ \frac{\text{x}}{\text{a}}=\sec\theta\cos\phi\ .....\text{i}$
$\text{y}=\text{b}\sec\theta\sin\phi$
$\Rightarrow \frac{\text{y}}{\text{b}}=\sec\theta\sin\theta \ .....(\text{ii})$
$\Rightarrow\ \frac{\text{z}}{\text{c}}=\tan\theta\ .....\text{(iii)}$
We have to prove that $\frac{\text{x}^2}{\text{a}^2}+\frac{\text{y}^2}{\text{b}^2}-\frac{\text{z}^2}{\text{c}^2}=1$
Squaring the above equations and then subtracting the third from the sum of the first two, we have
$\Big(\frac{\text{x}}{\text{a}}\Big)^2+\Big(\frac{\text{y}}{\text{b}}\Big)^2-\Big(\frac{\text{z}}{\text{c}}\Big)^2=(\sec\theta\cos\phi)^2+(\sec\theta\sin\theta)-(\tan\theta)^2$
$\Rightarrow\ \frac{\text{x}^2}{\text{a}^2}+\frac{\text{y}^2}{\text{b}^2}+\frac{\text{z}^2}{\text{c}^2}=\sec^2\theta\cdot\cos^2\phi+\sec^2\theta\sin^2\phi-\tan^2\theta$
$\Rightarrow\ \frac{\text{x}^2}{\text{a}^2}+\frac{\text{y}^2}{\text{b}^2}+\frac{\text{z}^2}{\text{c}^2}=(\sec^2\cos^2\phi+\sec^2\theta\sin^2\phi)-\tan^2\theta$
$\Rightarrow\ \frac{\text{x}^2}{\text{a}^2}+\frac{\text{y}^2}{\text{b}^2}+\frac{\text{z}^2}{\text{c}^2}=\sec^2\theta(\cos^2\phi+\sin^2\phi)-\tan^2\theta$
$\Rightarrow\ \frac{\text{x}^2}{\text{a}^2}+\frac{\text{y}^2}{\text{b}^2}+\frac{\text{z}^2}{\text{c}^2}=\sec^2\theta(1)-\tan^2\theta$
$\Rightarrow\ \frac{\text{x}^2}{\text{a}^2}+\frac{\text{y}^2}{\text{b}^2}+\frac{\text{z}^2}{\text{c}^2}=\sec^2\theta-\tan^2\theta$
$\Rightarrow\ \frac{\text{x}^2}{\text{a}^2}+\frac{\text{y}^2}{\text{b}^2}+\frac{\text{z}^2}{\text{c}^2}=1$
Hence proved.