Prove the following trigonometric identities- 1-tan2-1-cot2-1-tan-1-cot- 2-tan2-

We know that sec^2θ= 1 + tan^2θ and cosec^2θ= 1 + cot^2θ secθ = 1/cosθ cosecθ = 1/sinθ LHS = 1 + tan^2θ/1 + cot^2θ = sec^2θ/cosec^2θ = (secθ/cosecθ)^2 = (1/c ...

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Standard X

Mathematics

Trigonometric Identities

Prove the fol...

Question

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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Solution

We know that $s e c^{2} θ = 1 + t a n^{2} θ$ and $c o s e c^{2} θ = 1 + c o t^{2} θ$
$s e c θ = \frac{1}{c o s θ}$
$c o s e c θ = \frac{1}{s i n θ}$

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

$= \frac{s e c^{2} θ}{c o s e c^{2} θ}$

$= (\frac{s e c θ}{c o s e c θ})^{2}$

$= (\frac{1}{c o s θ} \times s i n θ)^{2}$

$= (\frac{s i n θ}{c o s θ})^{2}$

$= (t a n θ)^{2}$

$= t a n^{2} θ = R H S$

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Prove the following trigonometric identities: 1+tan2θ1+cot2θ=(1−tan θ1−cot θ)2=tan2θ

We know that sec2θ=1+tan2θ and cosec2θ=1+cot2θ secθ=1cosθ cosecθ=1sinθ LHS=1+tan2θ1+cot2θ =sec2θcosec2θ =(secθcosecθ)2 =(1cosθ×sinθ)2 =(sinθcosθ)2 =(tanθ)2 =tan2θ =RHS

Prove the following trigonometric identities:

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