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Telangana TET Paper 2 Math and Science Model Paper 4

Section: Math and Science

Question : 123 of 150

Marks: +1, -0

\frac{[(sinx+siny)(siny-sinx)]}{[(cosx+cosy)(cosy-cosx)]}

Solution:

Concept:

Use the identity

(a+b)(a-b)=a^2-b^2

and

\sin^2\theta+\cos^2\theta=1

to simplify fractions.

Explanation:

First, apply the difference of squares to numerator and denominator:

Numerator:

(\sin x+\sin y)(\sin y-\sin x)=\sin^2y-\sin^2x

Denominator:

(\cos x+\cos y)(\cos y-\cos x)=\cos^2y-\cos^2x

Now rewrite

\cos^2y=1-\sin^2y

and

\cos^2x=1-\sin^2x

Denominator becomes

(1-\sin^2y)-(1-\sin^2x)=\sin^2x-\sin^2y = -(\sin^2y-\sin^2x)

Thus the whole fraction is

\dfrac{\sin^2y-\sin^2x}{-(\sin^2y-\sin^2x)}=-1

Answer:

-1

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