Question

If 2tanθ=3 , thensin2θ+cos2θ=​

Answer 1

Answer:

=cosθ(Given)

=cosθ(Given)⇒θ=

=cosθ(Given)⇒θ= 4

=cosθ(Given)⇒θ= 4π

=cosθ(Given)⇒θ= 4π

=cosθ(Given)⇒θ= 4π

=cosθ(Given)⇒θ= 4π Now,

=cosθ(Given)⇒θ= 4π Now,2tanθ+cos

=cosθ(Given)⇒θ= 4π Now,2tanθ+cos 2

=cosθ(Given)⇒θ= 4π Now,2tanθ+cos 2 θ

=cosθ(Given)⇒θ= 4π Now,2tanθ+cos 2 θ=2tan

=cosθ(Given)⇒θ= 4π Now,2tanθ+cos 2 θ=2tan 4

=cosθ(Given)⇒θ= 4π Now,2tanθ+cos 2 θ=2tan 4π

=cosθ(Given)⇒θ= 4π Now,2tanθ+cos 2 θ=2tan 4π

=cosθ(Given)⇒θ= 4π Now,2tanθ+cos 2 θ=2tan 4π +cos

=cosθ(Given)⇒θ= 4π Now,2tanθ+cos 2 θ=2tan 4π +cos 2

=cosθ(Given)⇒θ= 4π Now,2tanθ+cos 2 θ=2tan 4π +cos 2

=cosθ(Given)⇒θ= 4π Now,2tanθ+cos 2 θ=2tan 4π +cos 2 4

=cosθ(Given)⇒θ= 4π Now,2tanθ+cos 2 θ=2tan 4π +cos 2 4π

=cosθ(Given)⇒θ= 4π Now,2tanθ+cos 2 θ=2tan 4π +cos 2 4π

=cosθ(Given)⇒θ= 4π Now,2tanθ+cos 2 θ=2tan 4π +cos 2 4π

=cosθ(Given)⇒θ= 4π Now,2tanθ+cos 2 θ=2tan 4π +cos 2 4π =2(1)+(

=cosθ(Given)⇒θ= 4π Now,2tanθ+cos 2 θ=2tan 4π +cos 2 4π =2(1)+( 2

=cosθ(Given)⇒θ= 4π Now,2tanθ+cos 2 θ=2tan 4π +cos 2 4π =2(1)+( 2

=cosθ(Given)⇒θ= 4π Now,2tanθ+cos 2 θ=2tan 4π +cos 2 4π =2(1)+( 2

=cosθ(Given)⇒θ= 4π Now,2tanθ+cos 2 θ=2tan 4π +cos 2 4π =2(1)+( 2 1

=cosθ(Given)⇒θ= 4π Now,2tanθ+cos 2 θ=2tan 4π +cos 2 4π =2(1)+( 2 1

=cosθ(Given)⇒θ= 4π Now,2tanθ+cos 2 θ=2tan 4π +cos 2 4π =2(1)+( 2 1 )

=cosθ(Given)⇒θ= 4π Now,2tanθ+cos 2 θ=2tan 4π +cos 2 4π =2(1)+( 2 1 ) 2

=cosθ(Given)⇒θ= 4π Now,2tanθ+cos 2 θ=2tan 4π +cos 2 4π =2(1)+( 2 1 ) 2

=cosθ(Given)⇒θ= 4π Now,2tanθ+cos 2 θ=2tan 4π +cos 2 4π =2(1)+( 2 1 ) 2 =2+

=cosθ(Given)⇒θ= 4π Now,2tanθ+cos 2 θ=2tan 4π +cos 2 4π =2(1)+( 2 1 ) 2 =2+ 2

=cosθ(Given)⇒θ= 4π Now,2tanθ+cos 2 θ=2tan 4π +cos 2 4π =2(1)+( 2 1 ) 2 =2+ 21

=cosθ(Given)⇒θ= 4π Now,2tanθ+cos 2 θ=2tan 4π +cos 2 4π =2(1)+( 2 1 ) 2 =2+ 21

=cosθ(Given)⇒θ= 4π Now,2tanθ+cos 2 θ=2tan 4π +cos 2 4π =2(1)+( 2 1 ) 2 =2+ 21

=cosθ(Given)⇒θ= 4π Now,2tanθ+cos 2 θ=2tan 4π +cos 2 4π =2(1)+( 2 1 ) 2 =2+ 21 =

=cosθ(Given)⇒θ= 4π Now,2tanθ+cos 2 θ=2tan 4π +cos 2 4π =2(1)+( 2 1 ) 2 =2+ 21 = 2

=cosθ(Given)⇒θ= 4π Now,2tanθ+cos 2 θ=2tan 4π +cos 2 4π =2(1)+( 2 1 ) 2 =2+ 21 = 25

=cosθ(Given)⇒θ= 4π Now,2tanθ+cos 2 θ=2tan 4π +cos 2 4π =2(1)+( 2 1 ) 2 =2+ 21 = 25