Question

plz answer this question guys ​

Answer 1

Simplify:

Step-by-step explanation:

$\ Given \ values:\\\\\sec^2 \pi + 2 \sin^2 \frac{3\pi}{2} - \tan^2\frac{5\pi}{3}\\\\\ Solution:\\\\\sec^2 \pi + 2 \sin^2 \frac{3\pi}{2} - \tan^2\frac{5\pi}{3}\\\\\therefore \pi =180^{\circ} \\\\\frac{3\pi}{2} = \frac{3\times 180^{\circ}}{2} \ =3\times 90^{\circ} = 270^{\circ} \\\\\frac{5\pi}{3} = \frac{5\times 180^{\circ}}{3} \ =5\times 60^{\circ} = 300^{\circ} \\\\\sec^2 180^{\circ} + 2 \sin^2 270^{\circ} - \tan^2 300^{\circ}$

$\therefore\\\\\sec180^{\circ}= -1\\\sin270^{\circ}=-1\\\tan300^{\circ}=\tan(270+300)=-\cot30^{\circ}=-\sqrt{3}$

$\ value: \\\\\sec^2 180^{\circ} + 2 \sin^2 270^{\circ} - \tan^2 300^{\circ}\\\\\ put \ the \ calculated \ value \ in \ above \ equation:\\\\\rightarrow (-1)^2+2\times(-1)^2-(-\sqrt{3})^2 \\\\\rightarrow 1+2\times 1 -(3) \\\\\rightarrow 1+2 -3 \\\\\rightarrow 3-3\\\\\rightarrow 0$

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