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Relation between Inverses of Trigonometric Functions and Their Reciprocal Functions

Prove that si...

Question

Prove that ${sin}^{2} \frac{π}{6} + {cos}^{2} \frac{π}{3} - {tan}^{2} \frac{π}{4} = - \frac{1}{2}$

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Solution

L.H.S. $= {sin}^{2} \frac{π}{6} + {cos}^{2} \frac{π}{3} - {tan}^{2} \frac{π}{4}$
$= {(\frac{1}{2})}^{2} + {(\frac{1}{2})}^{2} - (1)^{2}$
= $\frac{1}{4} + \frac{1}{4} - 1$
$= - \frac{1}{2} =$ R.H.S.
So, L.H.S. = R.H.S.
${sin}^{2} \frac{π}{6} + {cos}^{2} \frac{π}{3} - {tan}^{2} \frac{π}{4} = - \frac{1}{2}$
Hence proved.

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Relation between Inverses of Trigonometric Functions and Their Reciprocal Functions

Standard XII Mathematics

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