Tan2-1-tan2- -cot2-1-cot2-1

To prove, tan^2 θ/1+tan^2 θ + cot^2 θ/1+cot^2 θ = 1 LHS = tan^2 θ/1+tan^2 θ + cot^2 θ/1+cot^2 θ = tan^2 θ/sec^2 θ + cot^2 θ/cosec^2 θ = 1/sec^2 θ×sin^2 ...

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

Byju's Answer

Standard X

Mathematics

Trigonometric Identities

tan2θ/1+tan2θ...

Question

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

Open in App

Solution

$T o p r o v e, \frac{t a n^{2} θ}{1 + t a n^{2} θ} + \frac{c o t^{2} θ}{1 + c o t^{2} θ} = 1 L H S = \frac{t a n^{2} θ}{1 + t a n^{2} θ} + \frac{c o t^{2} θ}{1 + c o t^{2} θ} = \frac{t a n^{2} θ}{s e c^{2} θ} + \frac{c o t^{2} θ}{c o s e c^{2} θ} = \frac{1}{s e c^{2} θ} \times \frac{s i n^{2} θ}{c o s^{2} θ} + \frac{1}{c o s e c^{2} θ} \times \frac{c o s^{2} θ}{s i n^{2} θ} c o s^{2} θ \times \frac{s i n^{2} θ}{c o s^{2} θ} + s i n^{2} θ \times \frac{c o s^{2} θ}{s i n^{2} θ} = s i n^{2} θ + c o s^{2} θ = 1 = R H S$

Suggest Corrections

Similar questions

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

\frac{1 + {tan}^{2} θ}{1 + {cot}^{2} θ} =

(i) $(s e c^{2} θ - 1) c o t^{2} θ = 1$ (ii) $(s e c^{2} θ - 1) (c o s e c^{2} θ - 1) = 1$

(iii) $(1 - c o s^{2} θ) s e c^{2} θ = t a n^{2} θ$

Prove:

$(i) {sin}^{2} θ + \frac{1}{1 + {tan}^{2} θ} = 1$

$(i i) \frac{1}{1 + {tan}^{2} θ} + \frac{1}{1 + {cot}^{2} θ} = 1$

Q. Prove that:

{sec}^{2} θ = 1 + {tan}^{2} θ

Join BYJU'S Learning Program

tan2θ(1+tan2θ)+cot2θ(1+cot2θ)=1

To prove, tan2 θ1+tan2 θ+cot2 θ1+cot2 θ=1LHS=tan2 θ1+tan2 θ+cot2 θ1+cot2 θ=tan2 θsec2 θ+cot2 θcosec2 θ=1sec2 θ×sin2 θcos2 θ+1cosec2 θ×cos2 θsin2 θcos2 θ×sin2 θcos2 θ+sin2 θ×cos2 θsin2 θ=sin2 θ+cos2 θ=1=RHS

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