Questions
If $\tan 4x+\tan 5x-\tan 9x=k\tan 4x\tan 5x\tan 9x$ then $k=$
- $1$
- $-1$
- $ \pm 1$
- $2$
State true or false $\tan(\dfrac{\pi}{4} + \theta) - \tan(\dfrac{\pi}{4} -\theta) = 2\tan\theta$
- True
- False
State true or false
- True
- False
$tan 5x-tan 3x-tan 2x=$
- $\tan 5x \tan 3x \tan 2x$
- $\sin 5x \sin 3x \sin 2x$
- $\cos 5x \cos 3x \cos 2x$
- $\sec 5x \sec 3x \sec 2x$
$A, B, C$ are three angles such that $\tan A+\tan B+\tan C=\tan A \tan B \tan C.$ Which of the following statements is always correct ?
- $ABC$ is a triangle, i.e. $A+B+C=\pi $
- $A=B=C. i.e., $ $ABC$ is an equilateral triangle
- $A+B=C, $ i.e., $ABC$ is a right- angled triangle
- $A+B=\pi $
If $\dfrac{\pi}{4}<A<\dfrac{\pi}{2}$ then $\tan^{-1}\left(\dfrac{1}{2}\tan 2A\right)+\tan^{-1}(\cot A)+\tan^{-1}(\cot^{3}A)$=
- $0$
- $\pi$
- $\pi/2$
- $\pi/4$
If $A+B+C=\pi $ and cosA=cosB cosC, then tanB tanC is equal to
- $\frac { 1 }{ 2 } $
- $2$
- $1$
- $-\frac { 1 }{ 2 } $
$\alpha, \beta$ are the solution (s) of $3 cos 2 \theta + 4 sin 2 \theta = 5$
$tan (\alpha + \beta) = $
- $1$
- $\dfrac{3}{4}$
- $\dfrac{4}{3}$
- $\dfrac{1}{4}$
$\alpha, \beta$ are the solution (s) of $3 cos 2 \theta + 4 sin 2 \theta = 5$
$tan (\alpha - \beta) = $
- $0$
- $1$
- $\dfrac{1}{4}$
- $\dfrac{4}{3}$
In $\Delta$ ABC, (a + b + c) ( tan $\dfrac{A}{2}$ + tan $\dfrac{B}{2}$) =
- 2 c cot $\dfrac{A}{2}$
- 2 c cot $\dfrac{B}{2}$
- 2 c cot $\dfrac{C}{2}$
- 2 c tan $\dfrac{C}{2}$
$\cot^{2} \dfrac{\pi}{11}+\cot^{2} \dfrac{2\pi}{11}+\cot^{2} \dfrac{3\pi}{11}........+\cot^{2} \dfrac{5\pi}{11}=?$
- $15$
- $45$
- $9$
- $18$
$\tan \alpha + 2\tan 2\alpha + 4\tan 4\alpha + 8\tan 8\alpha + 16\tan 16\alpha + 32\cot 32\alpha $ is equal
- $\cot \alpha $
- $\tan \alpha $
- $\cos \alpha $
- $sin \alpha $
Simplify: $\tan 5^\circ \tan 30^\circ \times 4\tan 85^\circ$
- $1$
- $4$
- $4/\surd 3$
- $4\surd 3$
If $\alpha$ is the angle of first quadrant such that $co\sec ^{ 4 }{ \alpha }=17+\cot ^{ 4 }{ \alpha } $, then what is the value of $\sin{\alpha}$?
- $\cfrac{1}{3}$
- $\cfrac{1}{4}$
- $\cfrac{1}{9}$
- $\cfrac{1}{16}$
General solution of $\dfrac{1-{tan}^{2}x}{{sec}^{2}x}=\dfrac{1}{2}$ is
- $n\pi+\dfrac{\pi}{6},n\in Z$
- $n\pi-\dfrac{\pi}{6},n\in Z$
- $n\pi\pm\dfrac{\pi}{6},n\in Z$
- $2n\pi\pm\dfrac{\pi}{6},n\in Z$
The cosine of the obtuse angle formed by the medians from the vertices of the acute angles of an isosceles right angled triangle is
- $- 2 / 3$
- $- 4 / 5$
- $- 3 / 5$
- $- 3 / 4$
In an isosceles $\triangle ABC$, if the altitudes intersect on the inscribed circle then cosine of the vertical angle $'A'$ is :
- $\cfrac{1}{9}$
- $\cfrac{1}{3}$
- $\cfrac{2}{3}$
- None of these
If $3sin\alpha =5sin\beta ,\quad then\quad \frac { \tan { \frac { \alpha +\beta }{ 2 } } }{ \tan { \frac { \alpha -\beta }{ 2 } } } $ is equal to
- $1$
- $2$
- $3$
- $4$
If $y\tan (A+B+C)=x\tan (A+B-C)=\lambda$, then $\tan 2C=?$
- $\dfrac{\lambda(x+y)}{\lambda^2-xy}$
- $\dfrac{\lambda(x+y)}{\lambda^2+xy}$
- $\dfrac{\lambda(x-y)}{xy-\lambda^2}$
- $\dfrac{\lambda (x-y)}{xy+\lambda^2}$
If $4^{2, sin^2x}.16^{tan^2x}.2^{4, cos^2x} = 256 $ such that $0 < x < \dfrac{\pi}{2}$ then $x$ is equal to ___________.
- $\dfrac{\pi}{3}$
- $\dfrac{\pi}{4}$
- $\dfrac{\pi}{12}$
- $\dfrac{\pi}{24}$