Solve

Guides

0

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

Question

$3 - {sin}^{2} 48^{\circ} + {cos}^{2} 42^{\circ}$

Open in App

Solution

Verified by Toppr

$3 - {sin}^{2} 48^{\circ} + {cos}^{2} 42^{\circ}$
= $3 - {sin}^{2} (90 - 42)^{\circ} + {cos}^{2} 42^{\circ}$
= $3 - {cos}^{2} 42^{\circ} + {cos}^{2} 42^{\circ}$
= $3$

Was this answer helpful?

0

Similar Questions

Q1

3 - {sin}^{2} 48^{\circ} + {cos}^{2} 42^{\circ}

View Solution

Q2

{cos}^{2} 42^{\circ} - {sin}^{2} 48^{\circ}

View Solution

Q3

Find the value of

{sin}^{2} 48^{\circ} - {cos}^{2} 12^{\circ}

View Solution

Q4

Prove that

{sin}^{2} 12^{\circ} + {sin}^{2} 21^{\circ} + {sin}^{2} 39^{\circ} + {sin}^{2} 48^{\circ} - {sin}^{2} 9^{\circ} - {sin}^{2} 18^{\circ} = 1

.

View Solution

Q5

Without using trigonometric tables, prove that:
(i) cos 81° − sin 9° = 0
(ii) tan 71° − cot 19° = 0
(iii) cosec 60° − sec 30° = 0
(iv) cot 34° − tan56° = 0
(v) sin²48° + sin²42° = 1
(vi) cos²72° + cos²18° = 1

View Solution