Cot2 - cosec2 - - - 0 - - - 90

The correct option is A 1We know that, 1 + cot^2 θ = cosec^2 θ ⇒ cot^2 θ cosec^2 θ = 1 Hence, option A is correct. ...

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

Sum of Trigonometric Ratios in Terms of Their Product

cot2 - cosec2...

Question

$c o t^{2} - c o s e c^{2} θ =$ ............... $(0 < θ < 90)$

-1

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

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

Not defined

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

Open in App

Solution

Suggest Corrections

Similar questions

\sin θ = \frac{3}{4}

\sqrt{\frac{{cosec}^{2} θ - \cot^{2} θ}{\sec^{2} θ - 1}} = \frac{\sqrt{7}}{3}

Show that: $\frac{c o t (90^{\circ} - θ)}{c o s e c^{2} θ}$ . $\frac{s e c θ . c o t^{3} θ}{s i n^{2} (90^{\circ} - θ)}$ = $\sqrt{t a n^{2} θ + 1}$

Which of the following options is equal to the given expresssion?

$\frac{c o t (90^{\circ} - θ)}{c o s e c^{2} θ}$ $\times$ $\frac{s e c θ . c o t^{3} θ}{s i n^{2} (90^{\circ} - θ)}$

0 < θ < 90^{\circ}

s i n^{2} θ + c o s^{2} θ + t a n^{2} θ + c o t^{2} θ + s e c^{2} θ + + c o s e c^{2} θ

\frac{{cosec}^{2} θ - {cos}^{2} θ}{{cot}^{2} θ} = {sec}^{2} θ - {sin}^{2} θ

Join BYJU'S Learning Program