I sin2-1-1-tan2- 1 ii 1-1-tan2-1-1-cot2- -1

(i) sin^2θ+1/(1+tan^2θ)= 1 LHS = sin^2θ+1/(1+tan^2θ) =sin^2θ+1/(1+sin^2θ/cos^2θ) = sin^2θ+1/(cos^2θ + sin^2θ/cos^2θ) = sin^2θ+cos^2θ/cos^2θ + sin^2θ = ...

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Byju's Answer

Standard X

Mathematics

Trigonometric Identities

i sin2θ+1/1+t...

Question

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

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Solution

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

(ii) $\frac{1}{1 + t a n^{2} θ} + \frac{1}{1 + c o t^{2} θ} = 1 L H S = \frac{1}{1 + t a n^{2} θ} + \frac{1}{1 + c o t^{2} θ} = \frac{1}{1 + t a n^{2} θ} + \frac{1}{1 + \frac{1}{t a n^{2} θ}} = \frac{1}{1 + t a n^{2} θ} + \frac{1}{\frac{1 + t a n^{2} θ}{t a n^{2} θ}} = \frac{1}{1 + t a n^{2} θ} + \frac{t a n^{2} θ}{1 + t a n^{2} θ} = \frac{1 + t a n^{2} θ}{1 + t a n^{2} θ} = 1 = R H S$

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Prove:

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

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

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

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

Prove the following trigonometric identities:

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

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(i) sin2θ+1(1+tan2θ)=1 (ii) 11+tan2θ+11+cot2θ=1

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