Question

prove that sin^2π/6+cos^2π/3-tan^2π/4=-1/2​

Answer 1

Answer:

To prove:

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

LHS,

${ \sin}^{2} \frac{\pi}{6} + { \cos}^{2} \frac{\pi}{3} - { \tan}^{2} \frac{\pi}{4}$

${( \frac{1}{2} )}^{2} + {( \frac{1}{2} )}^{2} - {(1)}^{2}$

$\frac{1}{4} + \frac{1}{4} - 1$

$\frac{2}{4} - 1 = \frac{ - 2}{4}$

$\frac{ - 1}{2} = rhs$

Hence,Proved..

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Answer 2

Answer:

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Step-by-step explanation:

To prove:

{ \sin }^{2} \frac{\pi}{6} + { \cos}^{2} \frac{\pi}{3} - { \tan}^{2} \frac{\pi}{4} = \frac{ - 1}{2}sin26π+cos23π−tan24π=2−1

LHS,

{ \sin}^{2} \frac{\pi}{6} + { \cos}^{2} \frac{\pi}{3} - { \tan}^{2} \frac{\pi}{4}sin26π+cos23π−tan24π

{( \frac{1}{2} )}^{2} + {( \frac{1}{2} )}^{2} - {(1)}^{2}(21)2+(21)2−(1)2

\frac{1}{4} + \frac{1}{4} - 141+41−1

\frac{2}{4} - 1 = \frac{ - 2}{4}42−1=4−2

\frac{ - 1}{2} = rhs2−1=rhs

Hence,Proved..