Trigonometry

$$\sin \frac{{\pi x}}{2} = {x^2} - 2x + 2$$
Put value of x from options
$$\eqalign{ & x = 1$$
$$\sin \frac{\pi }{2} \times 1 = {1^2} - 2 \times 1 + 2$$
$$ ⇒ \sin {90^ \circ } = 1 - 2 + 2$$
$$ ⇒ 1 = 1$$

(secA - cosA)² + (cosecA - sinA)² - (cotA - tanA)²
= (sec²A + cos²A - 2secA.cosA) + (coses²A + sin²A - 2cosecA.sinA) - (cot²A + tan²A - 2cotA.tanA)
= sec²A - tan²A + cos²A + sin²A + coses²A - cot²A - 2
= 3 - 2
= 1
Alternate shortcut method:
(secA - cosA)² + (cosecA - sinA)² - (cotA - tanA)²
Put θ = 45°
= (sec45° - cos45°)² + (cosec45° - sin45°)² - (cot45° - tan45°)²
$$ = {\left( {\sqrt 2 - \frac{1}{{\sqrt 2 }}} \right)^2} + {\left( {\sqrt 2 - \frac{1}{{\sqrt 2 }}} \right)^2} - {\left( {1 - 1} \right)^2}$$
$$ = \frac{1}{2} + \frac{1}{2} - 0$$
$$ = 1$$

Details

$$\cos \pi x = {x^2} - x + \frac{5}{4}$$
$$ = {x^2} - 2 \times x \times \frac{1}{2} + \frac{1}{4} - \frac{1}{4} + \frac{5}{4}$$
$$ = {\left( {x - \frac{1}{2}} \right)^2} + 1 > 1$$
$$ = - 1 \leqslant \cos x \leqslant 1$$
∴ So, value of x is none of the above

sin²1° + sin²11° + sin²21° + sin²31° + sin²41° + sin²45° + sin²49° + sin²59° + sin²69° + sin²79° + sin²89°
= (sin²1° + sin²89°) + (sin²11° + sin²79°) + (sin²21° + sin²69°) + (sin²31° + sin²59°) + (sin²41° + sin²49°) + sin²45°
= 1 + 1 + 1 + 1 + 1 + $$\frac{1}{2}$$   [sin²A + sin²B = 1. If, A + B = 90°]
= $$5\frac{1}{2}$$

$${\sin ^2}\theta + {\cos ^2}\theta = 1$$
$${\text{Squaring both sides}}$$
$${\sin ^4}\theta + {\cos ^4}\theta $$
$$ = 1 - 2{\sin ^2}\theta . {\cos ^2}\theta $$
$${\text{Put}}\,\theta = {90^ \circ }$$
$$ = 1 - 2{\sin ^2}{90^ \circ } \times {\cos ^2}{90^ \circ }$$
$$ = 1 - 0$$
$$ = 1$$

$$\left( {2\sin \theta + 3\cos \theta } \right)$$
$${\text{Maximum value of}}$$
$$a\sin \theta + b\cos \theta $$
$$ = \sqrt {{a^2} + {b^2}} $$
$$ = \sqrt {{2^2} + {3^2}} $$
$$ = \sqrt {4 + 9} $$
$$ = \sqrt {13} $$

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