The correct option is B We know that (x 1/x)^2 ≥ 0 ⇒ x^2 + nbsp; 1/x^2 2 ≥ 0 Put x = tanθ ⇒ tan^2 θ + cot^2 θ ≥ 2 . ...

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

Tangent, Cotangent, Secant, Cosecant in Terms of Sine and Cosine

tan 2θ+ 2θ is

Question

$t a n^{2} θ + c o t^{2} θ$ is

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!

None of these

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

Open in App

Solution

The correct option is A

We know that $(x - \frac{1}{x})^{2}$ $\geq$ 0 $\Rightarrow$ $x^{2}$ + $\frac{1}{x^{2}}$ - 2 $\geq$ 0

Put x = $t a n θ$ $\Rightarrow$ $t a n^{2} θ + c o t^{2} θ$ $\geq$ 2 .

Suggest Corrections

Similar questions

$\frac{t a n^{2} θ}{(1 + t a n^{2} θ)} + \frac{c o t^{2} θ}{(1 + c o t^{2} θ)} = 1$

{cot}^{2} θ - t a n^{2} θ = {csc}^{2} θ - {sec}^{2} θ

{tan}^{2} θ + {cot}^{2} θ + 2 = {sec}^{2} θ . {csc}^{2} θ

s e c^{2} θ + c o s e c^{2} θ = s e c^{2} θ c o s e c^{2} θ

t a n^{2} θ - s i n^{2} θ = t a n^{2} θ s i n^{2} θ

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

{tan}^{2} θ + {cot}^{2} θ + 2 = {sec}^{2} θ \times {csc}^{2} θ

Join BYJU'S Learning Program