Use app×
Join Bloom Tuition
One on One Online Tuition
JEE MAIN 2026 Crash Course
NEET 2026 Crash Course
CLASS 12 FOUNDATION COURSE
CLASS 10 FOUNDATION COURSE
CLASS 9 FOUNDATION COURSE
CLASS 8 FOUNDATION COURSE
0 votes
131 views
in Trigonometry by (100 points)
closed by

The value of \( \sin \left(\frac{1}{2} \cot ^{-1}\left(-\frac{3}{4}\right)\right)+\cos \left(\frac{1}{2} \cot ^{-1}\left(-\frac{3}{4}\right)\right) \) is/are equal to [MULTIPLE CHOICES CORRECT(B,C,D) seeking the solution]

(A) 1 

(B) \( \frac{3 \sqrt{2}}{\sqrt{10}} \) 

(C) \( \sqrt{2} \sin \left(\frac{1}{2} \cot ^{-1}\left(-\frac{3}{4}\right)+\cot ^{-1}(1)\right) \) 

(D) \( \sqrt{2} \sin \left(\pi-\tan ^{-1}(1)-\frac{1}{2} \tan ^{-1} \frac{4}{3}\right) \)

1 Answer

+1 vote
by (53.1k points)
selected by
 
Best answer

Correct option is (B) \( \frac {3\sqrt 2} {\sqrt {10}}\)

\(\cot^{-1} (\frac{-3}4) = \theta\)

\(\Rightarrow \cot = \frac{-3}4 = \frac bp\)

\(b = -3, p = 4\)

\(h^2 = p^2 + b^2\)

\(= 4^2 + (-3)^2\)

\(= 16 + 9\)

\(= 25\)

\(h = 5\)

\(\cos \theta = \frac bh = \frac{-3}5\)

\(\sin \frac \theta 2 + \cos \frac \theta 2 \)

\(= \sqrt{\frac{1 - \cos \theta}2} + \sqrt {\frac {1 + \cos \theta}2}\)

\(= \sqrt{\frac {1 + \frac35}2} + \sqrt{\frac{1 - \frac 35}2}\)

\(=\sqrt{\frac8{10}} + \sqrt{\frac2{10}}\)

\(=\sqrt{\frac4{5}} + \sqrt{\frac1{5}}\)

\(= \frac 2 {\sqrt 5} + \frac 1{\sqrt 5}\)

\(= \frac 3 {\sqrt 5}\)

\(= \frac {3\sqrt 2} {\sqrt {10}}\)

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.

...