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Question 127
The value of $$\left[ \frac{\sin^2 24^\circ+\sin^2 66^\circ}{\cos^2 24^\circ+\cos^2 66^\circ} + \sin^2 61^\circ + \cos 61^\circ \sin 29^\circ \right]$$ is equal to:
Solution
we know that :
$$\sin^2\theta\ \ =\ \cos^2\left(90-\theta\ \right)$$
$$\sin\theta\ =\cos\left(90-\theta\ \right)$$
And $$\sin^2\theta\ +\cos^2\theta\ \ =1$$
Now using the above three identities
we get 1+sin^261+sin61cos29 = 1+$$1+\sin^261+\cos^261\ =2$$
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