Skip to main content
Back to INVERSE TRIGONOMETRY

JEE MATHEMATICS MCQ: INVERSE TRIGONOMETRY

1 Easy
1 of 1
Analyzing +4 | -1
If x1,x2,  x3,  x4{x_1},{x_2},\;{x_3},\;{x_4} are the roots of the equation x4−x3sin⁡2β−xcos⁡β−sin⁡β=0,  {x^4} - {x^3}\sin 2\mathrm{\beta } - x\cos \mathrm{\beta } - \sin \mathrm{\beta } = 0,\; then tan⁡−1+tan⁡−1x2+tan⁡−1x3+tan⁡−1x4{\tan ^{ - 1}} + {\tan ^{ - 1}}{x_2} + {\tan ^{ - 1}}{x_3} + {\tan ^{ - 1}}{x_4} is equal to
View answer and explanation
Correct option: B
π2−β\frac{\pi }{2} - \mathrm{\beta }
Option B (π2−β\frac{\pi }{2} - \mathrm{\beta }) is correct because by using Vieta's formulas and the arctangent addition formula, we can simplify the expression to π2−β\frac{\pi }{2} - \mathrm{\beta }.

More Questions from INVERSE TRIGONOMETRY

Practice All Chapter MCQs →
The value of cos⁡−1(cos⁡12)−sin⁡−1(sin⁡12)\cos ^{-1}(\cos 12)-\sin ^{-1}(\sin 12) is- The value of sin⁡(2tan⁡−1(1/3))+\sin \left(2 \tan ^{-1}(1 / 3)\right)+ cos⁡(tan⁡−122)\cos \left(\tan ^{-1} 2 \sqrt{2}\right) is Find cos⁡−1{12(cos⁡9π10−sin⁡9π10)}=\cos ^ {-1}\left\{\frac{1}{\sqrt{2}}\left(\cos \frac{9 \pi}{10}-\sin \frac{9 \pi}{10}\right)\right\}= If sin⁡(sin⁡−115+cos⁡−1x)=1\sin \left(\sin ^{-1} \frac{1}{5}+\cos ^{-1} x\right)=1, then xx is equal to If x2+y2+z2=r2,{x^2} + {y^2} + {z^2} = {r^2}, then

tan⁡−1(xyzr)+tan⁡−1(yzxr)+tan⁡−1(xzyr)  {\tan ^{ - 1}}\left( {\frac{{xy}}{{zr}}} \right) + {\tan ^{ - 1}}\left( {\frac{{yz}}{{xr}}} \right) + {\tan ^{ - 1}}\left( {\frac{{xz}}{{yr}}} \right)\; is equal to 2tan⁡−1(13)+tan⁡−1(17)2{\tan ^{ - 1}}\left( {\frac{1}{3}} \right) + {\tan ^{ - 1}}\left( {\frac{1}{7}} \right) is equal to
Type at least 2 characters or jump directly:
🩺 NEET MCQs 🔬 JEE MCQs 🎯 JEE College Predictor Study guides 📊 Dashboard
🔓

Save your learning progress

Sign in with a free account to save your practice answers and quiz results, then return to them when you revise.

Continue with Free Account →