Exercise \( 4.3 \)Evaluate the following:1. \( \sin \left[\cos ^{-1}\l

1.	Exercise \( 4.3 \)Evaluate the following:1. \( \sin \left[\cos ^{-1}\left(\frac{5}{13}\right)\right] \).2. \( \cos \left[\sin ^{-1}\left(-\frac{7}{25}\right)\right] \).3. \( \tan \left[\sin ^{-1}\left(\frac{15}{17}\right)\right] \).4. \( \cot \left[\operatorname{cosec}^{-1}\left(\frac{25}{7}\right)\right] \).5. \( \sin \left[\tan ^{-1}\left(\frac{3}{4}\right)\right] \).6. \( \sec \left[\cot ^{-1}\left(-\frac{5}{12}\right)\right] \).Answers \( 4.3 \)1. \( \frac{12}{13} \)2. \( \frac{24}{25} \)3. \( \frac{15}{8} \)4. \( \frac{24}{7} \)5. \( \frac{3}{5} \)6. \( \frac{13}{5} \)
Answer» (i) sin (cos^-1(\(\frac{5}{13}\))) = \(\sqrt{1-cos^2(cos^{-1}(\frac5{13}))}\) = \(\sqrt{1-(\frac5{13})^2}\) = \(\sqrt{\frac{13^2-5^2}{13^2}}\) = \(\frac{\sqrt{169-25}}{13}\) = \(\frac{\sqrt{144}}{13}\) = \(\frac{12}{13}\) (ii) cos(sin^-1(-7/25)) = \(\sqrt{1-sin^2(sin^-1(-7/25))}\) = \(\sqrt{1-(-7/25)^2}\) = \(\frac {\sqrt{25^2-7^2}}{25}\) = \(\frac{\sqrt{625-49}}{25}\) = \(\frac{\sqrt{576}}{25}\) = \(\frac{24}{25}\) (iii) tan(sin^-1(\(\frac{15}{17}\))) = \(\cfrac{sin(sin^{-1}(\frac{15}{17}))}{cos(sin^{-1}(\frac{15}{17}))}\) = \(\cfrac{\frac{15}{17}}{\sqrt{1-sin^2(sin^{-1}(\frac{15}{17}))}}\) = \(\cfrac{\frac{15}{17}}{\sqrt{1-\frac{15^2}{17^2}}}\) = \(\cfrac{\frac{15}{17}}{\frac{\sqrt{289-225}}{17}}\) = \(\frac{15}{\sqrt{64}}\) = \(\frac{15}8\) (iv) cot[cosec^-1(25/7)] = \(\sqrt{cosec^2(cosec^{-1}(25/7))-1}\) = \(\sqrt{(\frac{25}7)^2-1}\) = \(\sqrt{\frac{625-49}{7^2}}=\frac{\sqrt{576}}7=\frac{24}7\) (v) sin (tan^-13/4) = \(\frac{tan(tan^{-1}3/4)}{\sqrt{1+tan^2(tan^{-1}3/4)}}\) = \(\frac{3/4}{\sqrt{1+(3/4)^2}}\) = \(\cfrac{\frac34}{\frac{\sqrt{16+9}}4}\) = 3/5 (vi) sec (cot^-1(-5/12)) = \(\sqrt{1+\frac{1}{cot^2(cot(-5/12))}}\) = \(\sqrt{1+\frac{1}{(-5/12)^2}}\) = \(\sqrt{1+\frac{12^2}{5^2}}\) = \(\frac{\sqrt{25+144}}{5}\) = \(\frac{13}5\)

Exercise \( 4.3 \)Evaluate the following:1. \( \sin \left[\cos ^{-1}\left(\frac{5}{13}\right)\right] \).2. \( \cos \left[\sin ^{-1}\left(-\frac{7}{25}\right)\right] \).3. \( \tan \left[\sin ^{-1}\left(\frac{15}{17}\right)\right] \).4. \( \cot \left[\operatorname{cosec}^{-1}\left(\frac{25}{7}\right)\right] \).5. \( \sin \left[\tan ^{-1}\left(\frac{3}{4}\right)\right] \).6. \( \sec \left[\cot ^{-1}\left(-\frac{5}{12}\right)\right] \).Answers \( 4.3 \)1. \( \frac{12}{13} \)2. \( \frac{24}{25} \)3. \( \frac{15}{8} \)4. \( \frac{24}{7} \)5. \( \frac{3}{5} \)6. \( \frac{13}{5} \)

Answer»

(i) sin (cos^-1(\(\frac{5}{13}\))) = \(\sqrt{1-cos^2(cos^{-1}(\frac5{13}))}\)

= \(\sqrt{1-(\frac5{13})^2}\) = \(\sqrt{\frac{13^2-5^2}{13^2}}\) = \(\frac{\sqrt{169-25}}{13}\)

= \(\frac{\sqrt{144}}{13}\) = \(\frac{12}{13}\)

(ii) cos(sin^-1(-7/25)) = \(\sqrt{1-sin^2(sin^-1(-7/25))}\)

= \(\sqrt{1-(-7/25)^2}\)

= \(\frac {\sqrt{25^2-7^2}}{25}\) = \(\frac{\sqrt{625-49}}{25}\) = \(\frac{\sqrt{576}}{25}\) = \(\frac{24}{25}\)

(iii) tan(sin^-1(\(\frac{15}{17}\))) = \(\cfrac{sin(sin^{-1}(\frac{15}{17}))}{cos(sin^{-1}(\frac{15}{17}))}\) = \(\cfrac{\frac{15}{17}}{\sqrt{1-sin^2(sin^{-1}(\frac{15}{17}))}}\)

= \(\cfrac{\frac{15}{17}}{\sqrt{1-\frac{15^2}{17^2}}}\) = \(\cfrac{\frac{15}{17}}{\frac{\sqrt{289-225}}{17}}\) = \(\frac{15}{\sqrt{64}}\) = \(\frac{15}8\)

(iv) cot[cosec^-1(25/7)] = \(\sqrt{cosec^2(cosec^{-1}(25/7))-1}\)

= \(\sqrt{(\frac{25}7)^2-1}\) = \(\sqrt{\frac{625-49}{7^2}}=\frac{\sqrt{576}}7=\frac{24}7\)

(v) sin (tan^-13/4) = \(\frac{tan(tan^{-1}3/4)}{\sqrt{1+tan^2(tan^{-1}3/4)}}\)

= \(\frac{3/4}{\sqrt{1+(3/4)^2}}\)

= \(\cfrac{\frac34}{\frac{\sqrt{16+9}}4}\) = 3/5

(vi) sec (cot^-1(-5/12)) = \(\sqrt{1+\frac{1}{cot^2(cot(-5/12))}}\)

= \(\sqrt{1+\frac{1}{(-5/12)^2}}\)

= \(\sqrt{1+\frac{12^2}{5^2}}\) = \(\frac{\sqrt{25+144}}{5}\) = \(\frac{13}5\)

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