Prove that tan square A- sin square A=tan squareA.sin
square A
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Answered by
6
Solution:
LHS = tan²A - sin²A
= ( sin²A/cos²A ) - sin²A
= sin²A [ 1/cos²A - 1 ]
= sin²A( sec²A - 1 )
= sin²A tan²A
[ since , sec²A - 1 = tan²A ]
= tan²Asin²A
= RHS
•••••
LHS = tan²A - sin²A
= ( sin²A/cos²A ) - sin²A
= sin²A [ 1/cos²A - 1 ]
= sin²A( sec²A - 1 )
= sin²A tan²A
[ since , sec²A - 1 = tan²A ]
= tan²Asin²A
= RHS
•••••
Answered by
1
\sin^2A\cos^2B-\cos^2A\sin^2B=\sin^2A-\sin^2B proved.
Step-by-step explanation:
Consider the provided information.
\sin^2A\cos^2B-\cos^2A\sin^2B=\sin^2A-\sin^2B
Consider the LHS.
\sin^2A\cos^2B-\cos^2A\sin^2B
\sin^2A(1-\sin^2B)-(1-\sin^2A)\sin^2B (∴\cos^2x=1-\sin^2x)
\sin^2A-\sin^2A\sin^2B-\sin^2B+\sin^2A\sin^2B
\sin^2A-\sin^2B
Hence, proved.
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