SSC संयुक्त स्नातक स्तरीय टीयर-1 प्रैक्टिस सेट 1

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Question : 71 of 100

Marks: +1, -0

यदि

\sin ^{8} \θ+\cos ^{8} \θ-1=0

हो, तो

\cos ^{2} \θ\sin ^{2} \θ

का मान क्या है (यदि

\θ\≠0

या

\π\/ 2) ?

$-1$
0
1
2

Solution:

\sin ^{8} \θ+\cos ^{8} \θ-1=0

\⇒(\sin ^{4} \θ)^{2}+(\cos ^{4} \θ)^{2}+2 \sin ^{4} \θ

\.\cos ^{4} \θ-2 \sin ^{4} \θ\.\cos ^{4} \θ-1=0

\⇒(\sin ^{4} \θ+\cos ^{4} \θ)^{2}-2 \sin ^{4} \θ

\.\cos ^{4} \θ-1=0

\⇒[(\sin ^{2} \θ)^{2}+(\cos ^{2} \θ)^{2}+2 \sin ^{2} \θ

\cos ^{2} \θ-2 \sin ^{2} \θ\.\cos ^{2} \θ]^{2}

-2 \sin ^{4} \θ\.\cos ^{4} \θ-1=0

\⇒[(\sin ^{2} \θ+\cos ^{2} \θ)^{2}-2 \sin ^{2} \θ

.\cos ^{2} \θ]^{2}-2 \sin ^{4} \θ\. \cos ^{4} \θ-1=0

\∵\sin ^{2} \θ+\cos ^{2} \θ=1

\∴[1-2 \sin ^{2} \θ\.\cos ^{2} \θ]^{2}

-2 \sin ^{4} \θ\.\cos ^{4} \θ-1=0

\⇒[1+4 \sin ^{4} \θ \.\cos ^{4} \θ-2

× 2 \sin ^{2} \θ\. \cos ^{2} \θ-2 \sin ^{4} \θ

\.\cos ^{4} \θ-1=0

\⇒ 2 \sin ^{4} \θ\.\cos ^{4} \θ-4 \sin ^{2} \θ

\.\cos ^{2} \θ=0

⇒2 \sin ^{2} \θ\.\cos ^{2} \θ

(\sin ^{2} \θ\.\cos ^{2} \θ-2)=0

\⇒2 \sin ^{2} \θ\.\cos ^{2} \θ=0

\∴\sin ^{2} \θ\.\cos ^{2} \θ=0

(असत्य)

\∵\θ \≠0

अथवा

90°

\⇒\sin ^{2} \θ\.\cos ^{2} \θ-2=0

\⇒\sin ^{2} \θ\.\cos ^{2} \θ=2

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