EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR MATHEMATICS>Chapter 17 - Inverse Trigonometric Functions>COMEDK UGET 2014

Embibe Experts Solutions for Chapter: Inverse Trigonometric Functions, Exercise 1: COMEDK UGET 2015

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Embibe Experts Mathematics Solutions for Exercise - Embibe Experts Solutions for Chapter: Inverse Trigonometric Functions, Exercise 1: COMEDK UGET 2015

Attempt the free practice questions on Chapter 17: Inverse Trigonometric Functions, Exercise 1: COMEDK UGET 2015 with hints and solutions to strengthen your understanding. EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR MATHEMATICS solutions are prepared by Experienced Embibe Experts.

Questions from Embibe Experts Solutions for Chapter: Inverse Trigonometric Functions, Exercise 1: COMEDK UGET 2015 with Hints & Solutions

MEDIUM

COMEDK UGET

IMPORTANT

EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR MATHEMATICS>Chapter 17 - Inverse Trigonometric Functions>COMEDK UGET 2015>Q 1

$\cot^{- 1} (21) + \cot^{- 1} (13) + \cot^{- 1} (- 8) =$

EASY

COMEDK UGET

IMPORTANT

EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR MATHEMATICS>Chapter 17 - Inverse Trigonometric Functions>COMEDK UGET 2015>Q 1

$\tan (\cos^{- 1} (\frac{1}{5 \sqrt{2}}) - \sin^{- 1} (\frac{4}{\sqrt{17}}))$ is

HARD

COMEDK UGET

IMPORTANT

EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR MATHEMATICS>Chapter 17 - Inverse Trigonometric Functions>COMEDK UGET 2014>Q 1

If $\cot^{- 1} \frac{1}{5} + \cot^{- 1} \frac{1}{3} - \cot^{- 1} \frac{4}{7} = \cot^{- 1} x$ , then $x$ is equal to