JEE Main 2021 — Mathematics Trigonometry

$${({\mathrm{sin}}^{-1}x)}^{2}-{({\mathrm{cos}}^{-1}x)}^{2}=a$$(sin−1x)2−(cos−1x)2=a

Let ${\mathrm{sin}}^{-1}(x)=t$sin−1(x)=t

$$x=\mathrm{sin}t$$x=sint

$${({\mathrm{sin}}^{-1}\mathrm{sin}t)}^{2}-{({\mathrm{cos}}^{-1}\mathrm{sin}t)}^{2}=a$$(sin−1sint)2−(cos−1sint)2=a

$${t}^{2}-{(\frac{\pi }{2}-t)}^{2}=a$$t2−(2π​−t)2=a

$${t}^{2}-(\frac{{\pi }^{2}}{4}+{t}^{2}-\pi t)=a$$t2−(4π2​+t2−πt)=a

$$\pi t-\frac{{\pi }^{2}}{4}=a$$πt−4π2​=a

$$t=\frac{a}{\pi }+\frac{\pi }{4}$$t=πa​+4π​

$$x=\mathrm{sin}(\frac{a}{\pi }+\frac{\pi }{4})$$x=sin(πa​+4π​)

$$2{x}^{2}-1=2{\mathrm{sin}}^{2}(\frac{a}{\pi }+\frac{\pi }{4})-1$$2x2−1=2sin2(πa​+4π​)−1

$$=2{(\mathrm{sin}\frac{a}{\pi }\mathrm{cos}\frac{\pi }{4}+\mathrm{cos}\frac{a}{\pi }\mathrm{sin}\frac{\pi }{4})}^{2}-1$$=2(sinπa​cos4π​+cosπa​sin4π​)2−1

$$={(\mathrm{sin}\frac{a}{\pi }+\mathrm{cos}\frac{a}{\pi })}^{2}-1$$=(sinπa​+cosπa​)2−1

$$={\mathrm{sin}}^{2}\frac{a}{\pi }+{\mathrm{cos}}^{2}\frac{a}{\pi }+2\mathrm{cos}\frac{a}{\pi }\mathrm{sin}\frac{a}{\pi }-1$$=sin2πa​+cos2πa​+2cosπa​sinπa​−1

$$=2\mathrm{cos}\frac{a}{\pi }\mathrm{sin}\frac{a}{\pi }$$=2cosπa​sinπa​

$$=\mathrm{sin}\frac{2a}{\pi }$$=sinπ2a​