NTSE Stage 2 SAT 2016 Question Paper

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Question : 48 of 100

Marks: +1, -0

If

\cosec \x-\sin \x=\a

and

\sec \x-\cos \x=\b \x=\b

then:

$(\a^2\b)^{2/3}+(\ab^2)^{2/3}=1$
$(\a^2\b)^{2/3}+(\a^2\b^2)^{2/3}=1$
$\a^2+\b^2=1$
$\b^2-\a^2=1$

Solution:

\cosecx-\sin \x=\a;\sec \x-\cos \x=\b

\cosec \x-1/\cosecx=\a;\sec \x-1/\secx=\b

⇒{\cosecx^2\x-1}/\cosecx=\a;{\sec^2\x-1}/{\sec \x}=\b

⇒

{\cot^2\x}/{\cosec \x}=\a;{\tan^2 \x}/{\sec \x}=\b

\;\;\;\;{\cos^2\x}/{\sin \x}=\a;{\sin^2 \x}/{\cos \x}=\b

\;\;\;\;\;\a^2\b={\cos^4\x}/{\sin^2\x}.{\sin^2\x}/{\cos \x}=\cos^2 \x

⇒

\cos \x=(\a^2\b)^{1/2}

\;\;\;\;\cos^2\x=(\a^{2}\b)^{1/2}

Similarly,

\sin^2\x=(\ab^{2})^{2/3}

∴

\sin^{2}\x+\cos^2\x=1

⇒

(\ab^2)^{2/3}+(\a^2\b)^{2/3}=1

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