Prove that sin2-6- cos2-3- tan2-4-12

L.H.S.= sin^2 π6 + cos^2 π3 tan^2 π4 =(12)^2 +(12)^2 (1)^2 nbsp;= 14+14 1 = 12= R.H.S. So, L.H.S. = R.H.S. sin^2 π6+ cos^2 π3 tan^2 π4= 12 Hence proved. ...

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Standard XII

Mathematics

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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Similar questions

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Relation between Inverses of Trigonometric Functions and Their Reciprocal Functions

Standard XII Mathematics

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Prove that sin2π6+cos2π3−tan2π4=−12

L.H.S.=sin2π6+cos2π3−tan2π4 =(12)2+(12)2−(1)2 = 14+14−1 =−12= R.H.S. So, L.H.S. = R.H.S. sin2π6+cos2π3−tan2π4=−12 Hence proved.

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