Question

anyone plz help ull

full answee please​

Answer 1

Question

Prove that :

$\sf\dfrac{\cot\theta+\csc\theta}{\sin\theta+\tan\theta}=\cot\theta\times\csc\theta$

Solution :

We have to prove that :

$\rm\dfrac{\cot\theta+\csc\theta}{\sin\theta+\tan\theta}=\cot\theta\times\csc\theta$

Now ,If we convert LHS tanθ and cotθ in in terms of sinθ and cosθ, And Manipulate. We'll get answer.

Let's Try something Different :

First solve LHS

$\bf\blue{LHS}$

$\sf=\dfrac{\cot\theta+\csc\theta}{\sin\theta+\tan\theta}$

Multiply , Numerator and denominator by cotθ×cosec θ . Then ,

$\sf=\dfrac{\cot\theta+\csc\theta}{\sin\theta+\tan\theta}\times\dfrac{\csc\theta\times\cot\theta}{\csc\theta\times\cot\theta}$

$\sf=[\cot\theta\times\csc\theta]\times\dfrac{\cot\theta+\csc\theta}{\sin\theta+\tan\theta}\times\dfrac{1}{\csc\theta\times\cot\theta}$

$\sf=[\cot\theta\times\csc\theta]\times\dfrac{\cot\theta+\csc\theta}{(\sin\theta+\tan\theta)(\csc\theta\times\cot\theta)}$

$\sf=[\cot\theta\times\csc\theta]\times\dfrac{\cot\theta+\csc\theta}{\sin\theta\csc\theta\cot\theta+\tan\theta\cot\csc\theta}$

$\sf=[\cot\theta\times\csc\theta]\times\dfrac{\cot\theta+\csc\theta}{\cot\theta+\csc\theta}$

$\sf=\cot\theta\csc\theta$

$\bf\green{=RHS}$

Hence, Proved !

Answer 2

To prove,

(cosec θ - sin θ)(sec θ - cos θ)(tan θ + cot θ)=1

Proof:

LHS =(1/sin θ - sin θ)(1/cos θ - cos θ)(tan θ+1/tan θ)

=(1-sin²θ)/sinθ (1-cos²θ)/cosθ(1+tan²θ)/tanθ

=(cos²θ)/sinθ (sin²θ)/cosθ sec²θ/tanθ

=cos θ sin θ (1/cos²θ)/(sinθ/cosθ)

=cos θ sin θ(1/cos²θ)(cosθ/sinθ)

=cos θ sin θ (cos θ/sin θ cos²θ)

=cos θ sin θ (1/sin θ cos θ)

=cos θ sin θ/sin θ cos θ

=1=RHS

∴ Hence proved

Hope it Helps!!