Question

if tan A+sinA=m and A-sinA=n then show that m²-n²=4√mn​

Answer 1

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Step-by-step explanation:

LHS

• (tanA+sinA)²-(tanA-sinA)²
• 4tanAsinA. :- (a+b)² - (a-b)²= 4ab

now prove RHS

4√(tanA+sinA)(tanA-sinA)

=4√(tan²A-sin²A) √[sin²A/cos²A-sin²A]

=4√sin²A-sin²A cos²A÷cosA=4

sinA/cosA. √1-cos²A

=4tanA. √sin²A cos²A÷cosA=4•sinA/cosA

1-cos²A. ✓sin²A= 4tanA sinA.

(m²-n²)=4√mn thus LHS=RHS proved