Question : If $x=\sec 57^{\circ}$, then $\cot ^2 33^{\circ}+\sin ^2 57^{\circ}+\sin ^2 33^{\circ}+\operatorname{cosec}^2 57^{\circ} \cos ^2 33^{\circ}+\sec ^2 33^{\circ} \sin ^2 57^{\circ}$ is equal to:
Option 1: $2 x^2+1$
Option 2: $\frac{1}{x^2+1}$
Option 3: $x^2+1$
Option 4: $x^2+2$

Question

Question : The expression $\frac{\cos ^4 \theta-\sin ^4 \theta+2 \sin ^2 \theta+3}{(\operatorname{cosec} \theta+\cot \theta+1)(\operatorname{cosec} \theta-\cot \theta+1)-2}, 0^{\circ}<\theta<90^{\circ}$, is equal to:
Option 1: $\frac{1}{2} \sin \theta$
Option 2: $2 \sin \theta$
Option 3: $\sec \theta$
Option 4: $2\operatorname{cosec} \theta$

24 Views

Answer 1

Correct Answer: $x^2+2$

Solution :

Given, $x = \sec 57°$
$\sec θ = \frac{h}{b}$, $\tan θ = \frac{p}{b}$, $\sin θ = \frac{p}{h}$, $\operatorname{cosec} θ = \frac{h}{p}$, and $\cot θ = \frac{b}{p}$
Where p = perpendicular, b = base and h = hypotenuse
Also, $\sin θ × \operatorname{cosec} θ = 1$ and $\cos θ × \sec θ = 1$
When A + B = 90° then,
sin A = cos B, sec A = cosec B and tan A = cot B
$\sec 57°= x = \frac{h}{b}$
$p = \sqrt{x^2 - 1}$
$\cot^233° + \sin^2 57°+ \sin^233°+ \operatorname{cosec}^257° \cos^233°+ \sec^233° \sin^2 57°$
$=\cot^233° + \sin^2 57°+ \sin^2(90°-57°)+ \operatorname{cosec}^2(90°-33°) \cos^233°+ \sec^2(90°-57°) \sin^2 57°$
$=\cot^2 33°+ \sin^257°+ \cos^2 57° + \sec^2 33°\cos^2 33° + \operatorname{cosec}^2 57° \sin^2 57° $
$=\cot^2 33° + 1 + 1 + 1$
$=(\sqrt{\frac{x^2 - 1}{1}})^2 + 3$
$=x^2 - 1 + 3$
$=x^2 + 2$
Hence, the correct answer is $x^2 + 2$.