Question

if you answer this question I will mark you as brainliest ​

Answer 1

Answer:

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

Please mark it as brainliest answer if you like the answer and follow me ☺️☺️

Answer 2

Answer:the answer is 1