Prove the given trigonometric identity Class 10 From RD SHARMA EX~11.1 Q.No.23 (iii)
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▶ Question :-
Prove that :-
→ ( sinA - 2sin³A)/( 2cos³A - cosA ) = tan A .
Step-by-step explanation:
We have,
→ ( sinA - 2sin³A)/( 2cos³A - cosA ) = tan A .
==> ( sinA - 2sin³A) = ( 2cos³A - cos A ) tanA .
▶ Now, solving RHS .
= ( 2cos³A - cosA ) tanA .
= ( 2cos²A - 1 )cosA × sinA/cosA .
= [ 2( 1 - sin²A ) - 1 ] sinA .
= ( 2 - 2sin²A - 1 )sinA .
= ( 1 - 2sin²A ) sinA .
= sinA - 2sin³A = LHS .
•°• LHS = RHS .
Hence, it's proved .
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