RRB Junior Engineer Stage 1 Model Papers 8

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Question : 11 of 100

Marks: +1, -0

If

\sin^2α=\cos^3 α

then the value of

(\cot^6 α-\cot^2 α)

is

1
0
-1
2

Solution:

If

\sin^2α=\cos^3 α

\tan^2 α=\cos α

....(1)
Now consider

\cot^6 α-\cot^2 α

=1/{\tan^6 α}-1/{\tan^2 α}

Since

\cotα=1/{\tan α}

Substituting for

\tan^2 α

with

\cos α

from (1) above equation will be

=1/{cos^3 α}-1/{\cos α}={1-\cos^2 α}/{cos^3 α}={\sin ^2 α}/{\cos^3 α}={\tan^2 α}/{\cos α}=1

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