# Some Applications of Trigonometry - Class 10 - Mathematics

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## Extra Questions - Some Applications of Trigonometry | NCERT | Mathematics | Class 10

If $\sec 5A = \operatorname{cosec}(A-30^\circ)$, then the value of $A$ is

(A) $10^\circ$

(B) $20^\circ$

(C) $30^\circ$

(D) $15^\circ$

Given the equation: $$ \sec 5A = \operatorname{cosec}(A-30^\circ) $$

Using the identity **$\sec \theta = \operatorname{cosec}(90^\circ - \theta)$**, we can rewrite the left side as:
$$
\operatorname{cosec}(90^\circ - 5A)
$$

So the equation becomes: $$ \operatorname{cosec}(90^\circ - 5A) = \operatorname{cosec}(A - 30^\circ) $$

For the cosecant functions to be equal, their arguments must either be equal, or differ by a multiple of $360^\circ$ (period of the cosecant function). Since the arguments here are small, they must be equal directly: $$ 90^\circ - 5A = A - 30^\circ $$

Solving for $A$: $$ 90^\circ + 30^\circ = 5A + A \ 120^\circ = 6A \ A = \frac{120^\circ}{6} \ A = 20^\circ $$

Thus, the value of $A$ is **$20^\circ$**. Therefore, the correct option is (B) $\mathbf{20^\circ}$.

If $\cos \theta + \sin \theta = \sqrt{2} \sin \left(90^{\circ} - \theta\right)$, then $\cos \theta - \sin \theta =$

A $\sqrt{2} \sec \left(90^{\circ} - \theta\right)$

B $\sqrt{2} \sin \left(90^{\circ} - \theta\right)$

C $\sqrt{2} \sin \theta$

D $\sqrt{2} \cos \theta$

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If $y \sin \phi = x \sin (2 \theta + \phi)$, prove that $(x + y) \cot (\theta + \phi) = (y - x) \cot \theta$