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Trigonometric Identities — Mathematics STD 10 — Question

ICSE BoardEnglish MediumSTD 10MathematicsTrigonometric Identities2 Marks
Question
$
\text { If } \mathrm{x}=\mathrm{asec} \theta+\mathrm{b} \tan \theta \text { and } \mathrm{y}=\operatorname{atan} \theta+\mathrm{bsec} \theta \text {, prove that } x^2-y^2=a^2-b^2
$
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Answer

$
\begin{aligned}
& x^2-y^2=(a \sec \theta+b \operatorname{Tan} \theta)^2-(a \operatorname{Tan} \theta+b \operatorname{Sec} \theta)^2 \\
& \Rightarrow a^2 \sec ^2 \theta+b^2 \operatorname{Tan}^2 \theta+2 a b \operatorname{Sec} \theta \operatorname{Tan} \theta-\left(a^2 \operatorname{Tan}^2 \theta+b^2 \operatorname{Sec}^2 \theta+2 a b \operatorname{Sec} \theta \operatorname{Tan} \theta\right) \\
& \Rightarrow \sec ^2 \theta\left(a^2-b^2\right)+\operatorname{Tan}^2 \theta\left(b^2-a^2\right)=\left(a^2-b^2\right)\left[\operatorname{Sec}^2 \theta-\operatorname{Tan}^2 \theta\right] \\
& \Rightarrow\left(a^2-b^2\right) \quad\left[\text { Since } \sec ^2 \theta-\operatorname{Tan}^2 \theta=1\right]
\end{aligned}
$
Hence, $x^2-y^2=a^2-b^2$

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