Solve

Guides

0

You visited us 0 times! Enjoying our articles? Unlock Full Access!

Question

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

Open in App

Solution

Verified by Toppr

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

Was this answer helpful?

16

Similar Questions

Q1

Prove that:

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

View Solution

Q2

Prove that

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

.

View Solution

Q3

Show that

i)

sin \frac{π}{6} cos 0 + sin \frac{π}{4} cos \frac{π}{4} + sin \frac{π}{3} cos \frac{π}{6} = \frac{7}{4}

ii)

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

View Solution

Q4

The value of

{sin}^{2} \frac{π}{6} + {cos}^{2} \frac{π}{3} - {tan}^{2} \frac{π}{4} + {cot}^{2} \frac{π}{2}

is equal to

View Solution

Q5

Prove that

\frac{1 - {tan}^{2} (\frac{π}{4} - A)}{1 + {tan}^{2} (\frac{π}{4} - A)} = sin b A

Find b

View Solution