Question

If ɵ is an acute angle and sin ɵ = cos ɵ , find the value of 2 tan ² ɵ + sin ² ɵ -1.​

Answer 1

Answer:

sin©=cos©

tan©=1

2tan^2©+sin^2©-1

2+sin^2©-1

1+sin^2©

Answer 2

Given:

$\sf \to \sin \theta = \cos \theta$

Find:

$\sf \to 2{\tan}^{2} \theta + {\sin}^{2} \theta - 1$

Solution:

$\sf Let, \: denote \: \theta \: with \: A$

➟$\sf \sin A = \cos A$

➟$\sf \dfrac{\sin A}{\cos A} = 1$

$\sf we, \: know \: that \: \dfrac{\sin A}{\cos A} = \tan A$

$\sf Hence, \tan A = 1$

$\sf \to but \: we \: know \: that \tan {45}^{\circ} = 1$

$\sf Hence,A = {45}^{\circ}$

So,

$\sf \to 2{\tan}^{2} A + {\sin}^{2} A - 1$

$\sf \to 2{\tan}^{2} {45}^{\circ} + {\sin}^{2} {45}^{\circ} - 1$

where,

• tan² 45° = 1
• sin² 45° = 1/√2

So,

$\sf \to 2{\tan}^{2} {45}^{\circ} + {\sin}^{2} {45}^{\circ} - 1$

$\sf \to 2 \times {(1)}^{2} + {(\dfrac{1}{ \sqrt{2}})}^{2} - 1$

$\sf \to 2 \times 1 + \dfrac{1}{2} - 1$

$\sf \to 2 + \dfrac{1}{2} - 1$

$\sf \to \dfrac{4 + 1 - 2}{2}$

$\sf \to \dfrac{5 - 2}{2}$

$\sf \to \dfrac{3}{2}$

$\sf \therefore 2{\tan}^{2} \theta + {\sin}^{2} \theta - 1 = \dfrac{3}{2}$