Math, asked by 00987789, 3 days ago

(3-4sin^2θ) (sec^2θ-4tan^2θ)=(3-tan^2θ)(1-4sin^2θ)

Answers

Answered by bhaavyamedisetti3020
0

Answer:

LHS=  

cos  

2

θ

3−4sin  

2

θ

 

3sec  

2

θ−4tan  

2

θ

3(1+tan  

2

θ)−4tan  

2

θ

3+3tan  

2

θ−4tan  

2

θ

3−tan  

2

θ=RHS

Step-by-step explanation:

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