$\sec A(1-\sin A)(\sec A+\tan A)=?$
Solve it : $ \sqrt{\frac{\sec \theta-1}{\sec \theta+1}}+\sqrt{\frac{\sec \theta+1}{\sec \theta-1}} $
हल करें $ \sqrt{\frac{\sec \theta-1}{\sec \theta+1}}+\sqrt{\frac{\sec \theta+1}{\sec \theta-1}} \text { }$
The value of $\sin ^{2} 16^{\circ}-\cos ^{2} 74^{\circ}+1$ is:
$\sin ^{2} 16^{\circ}-\cos ^{2} 74^{\circ}+1$ का मान है:
Find the value of $\sin 190^{\circ}-\sin 185^{\circ}=?$
$\sin 190^{\circ}-\sin 185^{\circ}=?$ का मान ज्ञात कीजिए
If $\sec 3 x=\operatorname{cosec}\left(3 x-45^{\circ}\right)$, where $3 x$ is an acute angle, then $x$ is equal to
यदि $\sec 3 x=\operatorname{cosec}\left(3 x-45^{\circ}\right)$, जहाँ $3 x$ एक न्यून कोण है, तो $X$ बराबर है:
If $\cos (A-B)=\frac{\sqrt{3}}{2}$ and $\sec A=2,0^{\circ} \leq A \leq 90^{\circ}, 0^{\circ} \leq B \leq 90^{\circ}$ then what is the measure of $\mathrm{B}$ ?
यदि $\cos (A-B)=\frac{\sqrt{3}}{2}$ और $\sec A=2,0^{\circ} \leq A \leq 90^{\circ}, 0^{\circ} \leq B \leq 90^{\circ}$ है, तो $\mathrm{B}$ का माप क्या है ?
If $4 \sin ^2 \theta=3(1+\cos \theta), 0^{\circ}<\theta<90^{\circ}$, then what is the value of $(2 \tan \theta+4 \sin \theta-\sec \theta) ?$
यदि $4 \sin ^2 \theta=3(1+\cos \theta), 0^{\circ}<\theta<90^{\circ}$, तो $(2 \tan \theta+4 \sin \theta-\sec \theta)$ का मान क्या होगा?
The value of $\frac{\sin ^2 30^{\circ}+\cos ^2 60^{\circ}-\sec 35^{\circ} \cdot \sin 55^{\circ}}{\sec 60^{\circ}+\operatorname{cosec} 30^{\circ}}$ is equal to:
$\frac{\sin ^2 30^{\circ}+\cos ^2 60^{\circ}-\sec 35^{\circ} \cdot \sin 55^{\circ}}{\sec 60^{\circ}+\operatorname{cosec} 30^{\circ}}$ का मान बराबर है:
The value of $\frac{\sin 23^{\circ} \cos 67^{\circ}+\sec 52^{\circ} \sin 38^{\circ}+\cos 23^{\circ} \sin 67^{\circ}+\operatorname{cosec} 52^{\circ} \cos 38^{\circ}}{\operatorname{cosec}^2 20^{\circ}-\tan ^2 70^{\circ}}$ is
$\frac{\sin 23^{\circ} \cos 67^{\circ}+\sec 52^{\circ} \sin 38^{\circ}+\cos 23^{\circ} \sin 67^{\circ}+\operatorname{cosec} 52^{\circ} \cos 38^{\circ}}{\operatorname{cosec}^2 20^{\circ}-\tan ^2 70^{\circ}} $ का मान होगा
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