Question

if cosec theta - sin theta = l and sec theta - cos theta = m , then show that : l square m sq. (l sq. + m sq. +3)=1

Answer 1

Given :
Cosecθ-sinθ=l
Secθ-Cosθ=m

To prove that : l²m²(l²+m²+3)=1

Solution:
Cosecθ-sinθ=l
1/sinθ -sinθ=l
1-sin²θ/sinθ=l
cos²θ/sinθ=l [∵Sin²θ+Cos²θ=1]

Secθ-Cosθ=m
1/cosθ-cosθ=m
1-cos²θ/cosθ=m
sin²θ/cosθ=m [∵Sin²θ+Cos²θ=1]

⇒l²m²(l²+m²+3)
⇒(Cos²θ/sinθ)²(sin²θ/cosθ)²[(cos²θ/sinθ)²+(sin²θ/cosθ)²+3]

⇔sin²θCos²θ[Cos⁴θ/sin²θ +sin⁴θ/cos²θ +3]

=sin²θCos²θ[[Cos⁶θ+Sin⁶θ+3sin²θCos²θ/Sin²θCos²θ]

=(cos²θ)³ +( sin²θ)³ +3sin²θCos²θ

=    (Cos²  θ+sin²θ)³ - 3Sin²θCos²θ(Sin²θ+cos²θ] +3sin²θCos²θ
                                                [∵a³+b³=(a+b)³-3ab(a+b)]
=1-3Sin²θCos²θ(1)+3sin²θCos²θ
=1

Hence proved

Answer 2

hello mate

hope it help you