Question

If 2tanθ=3 , thensin2θ+cos2θ=​

Answer 1

Given : 2tanθ=3

To Find : sin2θ+cos2θ

Solution:

2tanθ=3

=> tanθ  = 3/2

=> Sec²θ = 1 + tan²θ

=> Sec²θ = 1 + (3/2)²

=> Sec²θ = 13/4

=>  cos²θ = 4/13  => cosθ =  ± 2/√13

Sin²θ +  cos²θ =  1

=> Sin²θ +  4/13  =  1

=> Sin²θ  = 9/13  => sinθ =  ± 3/√13

sin2θ+cos2θ

= 2sinθcosθ  + cos²θ - Sin²θ

= 2( ± 3/√13)(± 2/√13) +   4/13 - 9/13  

= 12/13  - 5/13

= 7/13

sin2θ+cos2θ = 7/13

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