Question

Square root of 2 sin^2 38°+3+2 sin ^52° / 3 cosec^2 17° -2- 3tan ^2 72°

Answer 1

Trigonometry: Trigonometry is the study of relations between angles and their ratios with various properties to find sine, cosine, tan, cosec, sec and cot ratios. Some properties can deduce angles in their general value apart from known definite ones. Now let us know some identities-

    1. sin²θ + cos²θ = 1

    2. sec²θ - tan²θ = 1

    3. cosec²θ - cot²θ = 1

    4. sin(90° - θ) = cosθ

    5. tan(90° - θ) = cotθ

Answer:

    Required value = √5

Step-by-step explanation:

Now, $\mathsf{\sqrt{\frac{2sin^{2}38^{\circ}+3+2sin^{2}52^{\circ}}{3cosec^{2}18^{\circ}-2-3tan^{2}72^{\circ}}}}$

$\mathsf{=\sqrt{\frac{2sin^{2}38^{\circ}+3+2\{sin(90^{\circ}-38^{\circ})\}^{2}}{3cosec^{2}18^{\circ}-2-3\{tan(90^{\circ}-18^{\circ})\}^{2}}}}$

$\mathsf{=\sqrt{\frac{2sin^{2}38^{\circ}+3+2cos^{2}38^{\circ}}{3cosec^{2}18^{\circ}-2-3cot^{2}18^{\circ}}}}$

$\mathsf{=\sqrt{\frac{2(sin^{2}38^{\circ}+cos^{2}38^{\circ})+3}{3(cosec^{2}18^{\circ}-cot^{2}18^{\circ})-2}}}$

$\mathsf{=\sqrt{\frac{2+3}{3-2}}}$

= √5

Answer 2

Answer: $\sqrt{5}$

Step-by-step explanation:

Given expression : $\sqrt{\mathsf{\frac{2sin^{2}38^{\circ}+3+2sin^{2}52^{\circ}}{3cosec^{2}18^{\circ}-2-3tan^{2}72^{\circ}}}}$

This will be equals to

$\sqrt{\mathsf{\frac{2sin^{2}38^{\circ}+3+2\{sin(90^{\circ}-38^{\circ})\}^{2}}{3cosec^{2}18^{\circ}-2-3\{tan(90^{\circ}-18^{\circ})\}^{2}}}}$

which becomes

$\sqrt{\frac{2sin^{2}38^{\circ}+3+2cos^{2}38^{\circ}}{3cosec^{2}18^{\circ}-2-3cot^{2}18^{\circ}}}$

Using identity :

$\sin^2 x+\cos^2x=1\\\\ cosec^2 x-\cot^2 x=1$

We have

$\sqrt{\frac{2(sin^{2}38^{\circ}+cos^{2}38^{\circ})+3}{3(cosec^{2}18^{\circ}-cot^{2}18^{\circ})-2}}\\\\=\sqrt{\frac{2+3}{3-2}}\\\\ =\sqrt{5}$