Question

please solve it I will mark them brainlist​

Answer 1

Answer:

l²m²(l²+m²+3)=1 is proved.

Step-by-step explanation:

Given :

Cosecθ-sinθ=l

Secθ-Cosθ=m

RTP:

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)

Hope it was clear and helpful.....