Prove that - sin 2 - - tan 2 - - cos 2 - - 2 -

sin^2θ+tan^2θ+cos^2θ #160; #160; #160; [∴sin^2θ+cos^2θ=1] ⇒ 1+tan^2θ=^2θ ...

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

Circular Measurement of Angle

Prove that : ...

Question

Prove that : ${sin}^{2} θ + {tan}^{2} θ + {cos}^{2} θ = {sec}^{2} θ$

Open in App

Solution

Suggest Corrections

Similar questions

\frac{{tan}^{2} θ}{{tan}^{2} θ - 1} + \frac{c o s e c^{2} θ}{{sec}^{2} θ - c o s e c^{2} θ} = \frac{1}{{sin}^{2} θ - {cos}^{2} θ}

\frac{(1 - \tan^{2} θ)}{(1 + \tan^{2} θ)} = ?

{sin}^{2} θ + {cos}^{2} θ + {sec}^{2} θ + c o s e c^{2} θ + {tan}^{2} θ + {cot}^{2} θ

\sqrt{1 + t a n^{2} θ} \sqrt{1 + c o t^{2} θ} \sqrt{1 - c o s^{2} θ} \sqrt{1 - s i n^{2} θ} =

{cos}^{2} θ - {sin}^{2} θ = {tan}^{2} α

{cos}^{2} α - {sin}^{2} α = {tan}^{2} θ

Join BYJU'S Learning Program