Question

If 0 < 0 < 90, 0, sinθ = cos30, then obtain the value of 2tan²θ – 1.

Answer 1

we know concepts :
cos(90–θ ) = sinθ, tan(90–θ ) = cotθ
sin(90–θ ) = cosθ , cot(90–θ ) = tanθ
cosec(90–θ ) = secθ ,sec(90–θ) = cosecθ

sinθ = cos30°
sinθ = $\frac{\sqrt{3}}{2}$
sinθ = sin60°
therefore,  θ = 60
Now, 2tan²θ – 1
= 2tan²60° – 1
= 2(√3)² – 1
=2 × 3 - 1
= 5

hence, 2tan²θ – 1 = 5

Answer 2

we know that:-
cos(90 - θ ) = sinθ,
tan(90 - θ ) = cotθ
sin(90 - θ ) = cosθ ,
cot(90 - θ ) = tanθ
cosec(90 - θ ) = secθ ,
sec(90 - θ) = cosecθ


sinθ = cos30°

sinθ = cos(90 - 60)°

sinθ = sin60°

therefore,  θ = 60

Now,

2tan²θ - 1

= 2tan²60° - 1

= 2(√3)² – 1

=2 * 3 - 1

= 5

hence, 2tan²θ – 1 = 5


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