Simplify: \(\dfrac{\sin^2A}{1-\cos A} + \dfrac{\sin^2A}{1+\cos A}\)
2
Combine the two fractions over a common denominator: \(\dfrac{\sin^2A(1+\cos A)+\sin^2A(1-\cos A)}{(1-\cos A)(1+\cos A)}\).
The numerator simplifies to \(\sin^2A\times2 = 2\sin^2A\), and the denominator to \(1-\cos^2A = \sin^2A\) (using the Pythagorean identity).
So the expression becomes \(\dfrac{2\sin^2A}{\sin^2A} = 2\).
Hence, the simplified value is 2.
If $\tan\theta = \frac{5}{12}$, $0 < \theta < \frac{\pi}{2}$, then the value of $\frac{\cos\theta + 5\cot\theta}{\text{cosec}\theta - \cos\theta}$ will be:
The value of 4 sin 230° + 3 cot 260° - 2 tan 245° is:
The value of 1 - sin 35° cos 55° is equal to:
If sin 3 θ = cos ( θ – 6°), then θ is:
If θ = 45°, then what will be the value of \(\frac{{\\sin \,\theta \, + \,\cos \,\theta }}{{\sin \,\theta \, - \,\cos \,\theta }}\) ?
If sin A = \(\frac{1}{2}\) and cos B = \(\frac{1}{2}\) then find A + B.