JEE Main 2023 — Mathematics Trigonometry

Given,

$${\mathrm{sin}}^{-1}(\frac{\alpha }{17})+{\mathrm{cos}}^{-1}(\frac{4}{5})-{\mathrm{tan}}^{-1}(\frac{77}{36})=0$$sin−1(17α​)+cos−1(54​)−tan−1(3677​)=0

$$\Rightarrow {\mathrm{sin}}^{-1}(\frac{\alpha }{17})={\mathrm{tan}}^{-1}(\frac{77}{36})-{\mathrm{cos}}^{-1}(\frac{4}{5})$$⇒sin−1(17α​)=tan−1(3677​)−cos−1(54​)

$$\Rightarrow {\mathrm{sin}}^{-1}(\frac{\alpha }{17})={\mathrm{tan}}^{-1}(\frac{77}{36})-{\mathrm{tan}}^{-1}(\frac{3}{4})$$⇒sin−1(17α​)=tan−1(3677​)−tan−1(43​)

$$\Rightarrow {\mathrm{sin}}^{-1}(\frac{\alpha }{17})={\mathrm{tan}}^{-1}(\frac{\frac{77}{36}-\frac{3}{4}}{1+\frac{77}{36}\times \frac{3}{4}})$$⇒sin−1(17α​)=tan−1(1+3677​×43​3677​−43​​)

$$\Rightarrow {\mathrm{sin}}^{-1}(\frac{\alpha }{17})={\mathrm{tan}}^{-1}(\frac{8}{15})$$⇒sin−1(17α​)=tan−1(158​)

$$\Rightarrow {\mathrm{sin}}^{-1}(\frac{\alpha }{17})={\mathrm{sin}}^{-1}(\frac{8}{17})$$⇒sin−1(17α​)=sin−1(178​)

$$\Rightarrow \alpha =8$$⇒α=8

Now solving ${\mathrm{sin}}^{-1}(\mathrm{sin}\alpha )+{\mathrm{cos}}^{-1}(\mathrm{cos}\alpha ) ={\mathrm{sin}}^{-1}(\mathrm{sin}\alpha )+{\mathrm{cos}}^{-1}(\mathrm{cos}\alpha )$sin−1(sinα)+cos−1(cosα)=sin−1(sinα)+cos−1(cosα)

$$\Rightarrow {\mathrm{sin}}^{-1}(\mathrm{sin}8)+{\mathrm{cos}}^{-1}(\mathrm{cos}8)=3\pi -8+8-2\pi =\pi$$⇒sin−1(sin8)+cos−1(cos8)=3π−8+8−2π=π