Skip to main content
Back to COMPOUND ANGLES

JEE MATHEMATICS MCQ: COMPOUND ANGLES

1 Easy
1 of 1
Analyzing +4 | -1
If m  tan⁡  (θ−30∘)=n  tan⁡  (θ+120∘)m\,\,\tan \,\,\left( {\theta - 30^\circ } \right) = n\,\,\tan \,\,\left( {\theta + 120^\circ } \right), then cos⁡  2θ\cos \,\,2\theta equals
View answer and explanation
Correct option: D
m+n2(m−n)\frac{{m + n}}{{2\left( {m - n} \right)}}
Option D (m+n2(m−n)\frac{{m + n}}{{2\left( {m - n} \right)}}) is correct because it is the result of simplifying the trigonometric equation and solving for cos⁡2θ\cos 2\theta.

More Questions from COMPOUND ANGLES

Practice All Chapter MCQs →
If tan⁡(A+B)=p\tan \left( {A + B} \right) = p and tan⁡(A−B)=q\tan \left( {A - B} \right) = q, then the value of tan⁡  2A\tan \,\,2A is The value of cos⁡  x  cos⁡  y  sin⁡  (x−y)+cos⁡  y  cos⁡  z  sin⁡  (y−z)\cos \,\,x\,\,\cos \,\,y\,\,\sin \,\,\left( {x - y} \right) + \cos \,\,y\,\,\cos \,\,z\,\,\sin \,\,\left( {y - z} \right)

+cos⁡  z  cos⁡  x  sin⁡  (z−x)+sin⁡  (x−y)sin⁡  (y−z)  sin⁡  (z−x)+ \cos \,\,z\,\,\cos \,\,x\,\,\sin \,\,\left( {z - x} \right) + \sin \,\,\left( {x - y} \right)\sin \,\,\left( {y - z} \right)\,\,\sin \,\,\left( {z - x} \right), is
The sides of a triangle are sin⁡α,cos⁡α\sin \alpha, \cos \alpha and 1+sin⁡αcos⁡α\sqrt{1+\sin \alpha \cos \alpha} for some 0<α<π20<\alpha<\frac{\pi}{2}. Then, the greatest angle of the triangle is If tan⁡  A=2  tan⁡  B+cot⁡  B,    then   2  tan⁡(A−B)  \tan \,\,A = 2\,\,\tan \,\,B + \cot \,\,B,\;\;\mathrm{then}\;\,2\,\,\tan \left( {A - B} \right)\mathrm{\;} is equal to

If A,B,CA,B,C are angles of a triangle, then the minimum value of tan⁡2  A2+tan⁡2  B2+tan⁡2  C2{\tan ^2}\,\,\frac{A}{2} + {\tan ^2}\,\,\frac{B}{2} + {\tan ^2}\,\,\frac{C}{2}, is If A>0,   B>0A > 0,\,\;B > 0 and A+B=π3A + B = \frac{\pi }{3}, then the maximum value of tan⁡A  tan⁡B\tan A\,\,\tan B, is

Type at least 2 characters or jump directly:
🔓

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.