Use app×
Join Bloom Tuition
One on One Online Tuition
JEE MAIN 2025 Foundation Course
NEET 2025 Foundation Course
CLASS 12 FOUNDATION COURSE
CLASS 10 FOUNDATION COURSE
CLASS 9 FOUNDATION COURSE
CLASS 8 FOUNDATION COURSE
0 votes
113 views
in Mathematics by (15 points)
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. \( \frac{13}{5} \)

Please log in or register to answer this question.

1 Answer

0 votes
by (44.2k points)

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

 = \(\sqrt{1+\frac{1}{(-5/12)^2}}\)

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

Welcome to Sarthaks eConnect: A unique platform where students can interact with teachers/experts/students to get solutions to their queries. Students (upto class 10+2) preparing for All Government Exams, CBSE Board Exam, ICSE Board Exam, State Board Exam, JEE (Mains+Advance) and NEET can ask questions from any subject and get quick answers by subject teachers/ experts/mentors/students.

Categories

...