Question

If tanA + sinA = m, tanA - sinA = n then show that m² - n² = ✓mn.

Answer 1

Answer:

Step-by-step explanation:

Arbazhaider Expert

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

Answer 2

Answer:

Step-by-step explanation:

tanθ+sinθ=m and tanθ-sinθ=n

∴, m²-n²

=(m+n)(m-n)

=(tanθ+sinθ+tanθ-sinθ)(tanθ+sinθ-tanθ+sinθ)

=(2tanθ)(2sinθ)

=4tanθsinθ

4√mn

=4√(tanθ+sinθ)(tanθ-sinθ)

=4√(tan²θ-sin²θ)

=4√{(sin²θ/cos²θ)-sin²θ}

=4√sin²θ{(1/cos²θ)-1}

=4sinθ√{(1-cos²θ)/cos²θ}

=4sinθ√(sin²θ/cos²θ)

=4sinθ√tan²θ

=4sinθtanθ

∴, LHS=RHS (Proved)