The value of cos - 2 -x -sin - 2 -x cos - 2 -x -sin - 2 -x 2 is

The correct option is C 1 ( cos( π #160; 2 +x ) #160; +sin( π #160; 2 +x ) #160; #160;cos( π #160; 2 x ) #160; sin( π #160; 2 ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Local Maxima

The value of ...

Question

The value of ${⎛ ⎜ ⎜ ⎜ ⎝ \frac{cos (\frac{π}{2} + x) + sin (\frac{π}{2} + x)}{cos (\frac{π}{2} - x) - sin (\frac{π}{2} - x)} ⎞ ⎟ ⎟ ⎟ ⎠}^{2}$ is

\frac{1 - sin 2 x}{1 + sin 2 x}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

\frac{1 + sin 2 x}{1 - sin 2 x}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

1

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

None of these

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

Suggest Corrections

Similar questions

∣ ∣ ∣ ∣ ∣ \begin{matrix} 1 + {sin}^{2} x & {cos}^{2} x & 4 sin 2 x {sin}^{2} x & 1 + {cos}^{2} x & 4 sin 2 x {sin}^{2} x & {cos}^{2} x & 1 + 4 sin 2 x \end{matrix} ∣ ∣ ∣ ∣ ∣ =

$\frac{1 - i}{1 + i}$ is equal to

\frac{d}{d x} \sqrt{\frac{1 - sin 2 x}{1 + sin 2 x}}

(0 < x < \frac{π}{2})

sin x + sin \frac{π}{8} \sqrt{(1 - cos x)^{2} + {sin}^{2} x} = 0

[\frac{5 π}{2}, \frac{7 π}{2}]

2 cos x \leq ∣ ∣ \sqrt{1 + sin 2 x} - \sqrt{1 - sin 2 x} ∣ ∣ \leq \sqrt{2}

x \in [0, 2 π]

[\frac{a π}{2}, \frac{b π}{2}]

a + b

Join BYJU'S Learning Program