Cos 68-cos 8-sin 68-sin 8-

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Byju's Answer

Standard XII

Mathematics

Evaluation of a Determinant

cos 68∘cos 8∘...

Question

cos68^o cos8^o + sin68^o sin8^o =?

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Solution

The correct option is A
½

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Similar questions

Q. Prove that:

(i)

\frac{\cos 9 ° + \sin 9 °}{\cos 9 ° - \sin 9 °} = \tan 54 °

(ii)

\frac{\cos 8 ° - \sin 8 °}{\cos 8 ° + \sin 8 °} = \tan 37 °

Q. Prove that
(i)

\frac{\cos 11 ° + \sin 11 °}{\cos 11 ° - \sin 11 °} = \tan 56 °

.

(ii)

\frac{\cos 9 ° + \sin 9 °}{\cos 9 ° - \sin 9 °} = \tan 54 °

(ii)

\frac{\cos 8 ° - \sin 8 °}{\cos 8 ° + \sin 8 °} = \tan 37 °

Prove that:
(i) $\frac{c o s 11^{\circ} + s i n 11^{\circ}}{c o s 11^{\circ} - s i n 11^{\circ}} = t a n 56^{\circ}$
(ii) $\frac{c o s 9^{\circ} + s i n 9^{\circ}}{c o s 9^{\circ} - s i n 9^{\circ}} = t a n 54^{\circ}$
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\int \frac{{sin}^{8} x - {cos}^{8} x}{1 - 2 {sin}^{2} x {cos}^{2} x} d x

\int \frac{{sin}^{8} x - {cos}^{8} x}{1 - 2 {sin}^{2} x {cos}^{2} x} d x

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cos68o cos8o + sin68o sin8o =?

The correct option is A ½

cos68^o cos8^o + sin68^o sin8^o =?

The correct option is A
½