Question

Given: cot32°=4/3 and cos44°=7/9 find sin62°

Answer 1

This question has the same problem as the other one you shared. Neither cot(32°) = 4/3 nor cos(44°) = 7/9 are true.

But if we indulge the author for a moment, we have from the double angle identity

$\sin(2x) = 2\sin(x) \cos(x)$

so if x = 32°, then

$\sin(64^\circ) = 2\sin(32^\circ) \cos(32^\circ)$

We also have from the Pythagorean identity

$\sin^2(x) + \cos^2(x) = 1 \implies 1 + \cot^2(x) = \csc^2(x) \\\\ \implies \csc(x) = \pm \sqrt{1+\cot^2(x)} \\\\ \implies \sin(x) = \pm \dfrac1{\sqrt{1+\cot^2(x)}}$

Now, 0° < 32° < 90°, so both sin(32°) and cos(32°) are positive.

If we assume cot(32°) = 4/3 to be true, it follows that

$\sin(32^\circ) = \dfrac1{\sqrt{1+\left(\frac43\right)^2}} = \dfrac35$

$\cos(32^\circ) = \sqrt{1 - \sin^2(32^\circ)} = \dfrac45$

$\implies \sin(64^\circ) = 2\times\dfrac35\times\dfrac45 = \dfrac{24}{25}$

so E is the only correct choice regardless of the baseless assumptions made by the question.