Trigonometry Practice Test 1

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Question : 1 of 40

Marks: +1, -0

If

{\cos x}/{\cos y}=n,{\sin x}/{\sin y}=m,

then what is

(m^2 - n^2 ) \sin^2 y

equal to?

$1 - n^2$
$1 + n^2$
$m^2$
$n^2$
None of these

Solution:

Given,

{\cos x}/{\cos y}=n,{\sin x}/{\sin y}=m

Now,

(m^2 - n^2 ) \sin^2 y

=({\sin^2x}/{\sin^2y}-{\cos^2x}/{\cos^2y})\sin^2y

={(1-\cos^2x)\cos^2y-\cos^2x(1-\cos^2y)}/{\cos^2y}

={\cos^2y-\cos^2x}/{\cos^2y}=1-n^2

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