Question

Prove that
1+cot theta/sin theta + 1+tan theta/cos theta = cosec theta sec² theta+sec theta cosec² theta
(using LHS/RHS method)



Spam=Report
Have a nice day ahead y'all!​

Answer 1

Step-by-step explanation:

Given :-

[(1+Cot θ)/Sin θ] + [ (1+ Tan θ)/Cos θ]

To find :-

Prove that

[(1+Cot θ)/Sin θ] + [ (1+ Tan θ)/Cos θ] = Cosec θ Sec² θ +Sec θ Cosec²θ

Solution :-

On taking LHS

[(1+Cot θ)/Sin θ] + [ (1+ Tan θ)/Cos θ]

=>[{1+(Cosθ/Sinθ)}/Sinθ]+

[{1+(Sinθ/Cosθ}/Cos θ]

=>[{(Sinθ+Cosθ)/Sinθ}/Sinθ]+

[{(Cosθ+Sinθ)/Cosθ}/Cosθ]

=> [(Sin θ+Cos θ)/Sin² θ] +

[ (Sin θ+Cos θ)/Cos²θ]

=> (Sin θ+Cos θ)[(1/Sin²θ)+(1/Cos²θ)]

=>(Sin θ + Cos θ)[(Cos² θ + Sin² θ)/

(Sin² θ Cos² θ)]

We know that

Sin² A + Cos² A = 1

=> (Sin θ+Cos θ) [1 /(Sin²θCos²θ)]

=> (Sin θ+Cos θ) /(Sin²θCos²θ)

=>[Sinθ/(Sin²θCos²θ)]+[Cosθ(Sin²θCos²θ)]

=> [1/(Sin θCos²θ)] + [ 1/(Sin² θCos θ)]

=> Cosec θ Sec² θ + Cosec²θ Sec θ

=> Cosec θ Sec² θ +Sec θ Cosec²θ

=> RHS

=> LHS = RHS

Hence, Proved.

Answer:-

[(1+Cot θ)/Sin θ] + [ (1+ Tan θ)/Cos θ]

= Cosec θ Sec² θ +Sec θ Cosec²θ

Used formulae:-

→ Sin² A + Cos² A = 1

→ Tan θ = Sin θ / Cos θ

→ Cot θ = Cos θ / Sin θ

→ 1/ Sin θ = Cosec θ

→ 1/ Cos θ = Sec θ