Is LHS-RHS- 1-1-1-1- - 2-sin2- - cos2-

The correct option is B Yes 1 cotθ1+cotθ+1+cotθ1 cotθ =sinθ cosθsinθ+cosθ+sinθ+cosθsinθ cosθ Simplifying, we get (sinθ+cosθ)^2+(sinθ cosθ)^2sin^2θ ...

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Byju's Answer

Standard XII

Physics

Chain Rule of Differentiation

Is LHS=RHS? ...

Question

Is LHS=RHS?
$\frac{1 + cot θ}{1 - cot θ} + \frac{1 - cot θ}{1 + cot θ} = \frac{2}{{sin}^{2} θ - {cos}^{2} θ}$

Yes

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Solution

The correct option is B Yes
$\frac{1 - c o t θ}{1 + c o t θ} + \frac{1 + c o t θ}{1 - c o t θ}$
$= \frac{s i n θ - c o s θ}{s i n θ + c o s θ} + \frac{s i n θ + c o s θ}{s i n θ - c o s θ}$
Simplifying, we get
$\frac{(s i n θ + c o s θ)^{2} + (s i n θ - c o s θ)^{2}}{s i n^{2} θ - c o s^{2} θ}$
$= \frac{2}{s i n^{2} θ - c o s^{2} θ}$

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Is LHS=RHS?1+cotθ1−cotθ+1−cotθ1+cotθ=2sin2θ−cos2θ

The correct option is B Yes1−cotθ1+cotθ+1+cotθ1−cotθ=sinθ−cosθsinθ+cosθ+sinθ+cosθsinθ−cosθSimplifying, we get(sinθ+cosθ)2+(sinθ−cosθ)2sin2θ−cos2θ=2sin2θ−cos2θ

Is LHS=RHS?
$\frac{1 + cot θ}{1 - cot θ} + \frac{1 - cot θ}{1 + cot θ} = \frac{2}{{sin}^{2} θ - {cos}^{2} θ}$