(A) Given that $\sin \theta = \cos 30$.We know that $\cos 30 = \frac{\sqrt{3}}{2}$.Therefore,$\sin \theta = \frac{\sqrt{3}}{2}$.Since $\sin 60 = \frac

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(A) Given that $\sin \theta = \cos 30$.We know that $\cos 30 = \frac{\sqrt{3}}{2}$.Therefore,$\sin \theta = \frac{\sqrt{3}}{2}$.Since $\sin 60 = \frac{\sqrt{3}}{2}$,we have $\theta = 60$.Now,substitute $\theta = 60$ into the expression $2 \tan^2 \theta - 1$:$2 \tan^2 60 - 1 = 2(\sqrt{3})^2 - 1$.$= 2(3) - 1 = 6 - 1 = 5$.

Class 10 Mathematics Introduction to Trigonometry Mix Examples - Introduction to Trigonometry

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If $0 < \theta < 90$ and $\sin \theta = \cos 30$,then $2 \tan^2 \theta - 1 = \dots$

A
$5$
B
$4$
C
$3$
D
$0$

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Mix Examples - Introduction to Trigonometry

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If $0 < \theta < 90$ and $\sin \theta = \cos 30$,then $2 \tan^2 \theta - 1 = \dots$

Similar Questions

$\frac{\cos ^{2} 40^{\circ}+\cos ^{2} 50^{\circ}}{\sin ^{2} 40^{\circ}+\sin ^{2} 50^{\circ}}=\ldots \ldots \ldots \ldots$

If $\tan A = \cot B$,then $A + B = \ldots$

If $\sin ^{2}(3 x+30^{\circ})+\cos ^{2}(2 x+45^{\circ})=1$,then $x = \dots$ (in $^{\circ}$)

In $\Delta ABC$,$m \angle C = 90^{\circ}$ and $\tan A = \frac{1}{\sqrt{3}}$,then $\sin A = \ldots$

Simplify $(1+\tan ^{2} \theta)(1-\sin \theta)(1+\sin \theta)$

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