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Trigonometric Identities — Maths STD 10 — Question

CBSE BoardEnglish MediumSTD 10MathsTrigonometric Identities5 Marks
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 $\frac{\text{x}^2}{\text{a}^2}+\frac{\text{y}^2}{\text{b}^2}-\frac{\text{z}^2}{\text{c}^2}=1.$
✓

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.

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