Question

prove that (cosec 0 - sin 0) . (sec 0 - cos 0) = 1/tan 0 + cot 0​

Answer 1

$(\csc\theta-\sin \theta)(\sec \theta - \cos \theta)=\dfrac{1}{\tan \theta + \cot\theta}$ Proved.

Step-by-step explanation:

Consider the provided expression.

$(\csc\theta-\sin \theta)(\sec \theta - \cos \theta)=\dfrac{1}{\tan \theta + \cot\theta}$

Consider the LHS.

$=(\csc\theta-\sin \theta)(\sec \theta - \cos \theta)\\=(\frac{1}{\sin\theta}-\sin \theta)(\frac{1}{\cos\theta} - \cos \theta)\\=(\frac{1-\sin^2\theta}{\sin\theta})(\frac{1- \cos^2 \theta}{\cos\theta})\\=(\frac{\cos^2\theta}{\sin\theta})(\frac{\sin^2 \theta}{\cos\theta})\\=\cos\theta\sin\theta$

Now Consider the RHS

$=\dfrac{1}{\tan \theta + \cot\theta}\\=\dfrac{1}{\frac{\sin\theta}{\cos\theta} +\frac{\cos\theta}{\sin\theta}}\\\\=\dfrac{\cos\theta\sin\theta}{{\sin^2\theta}+{\cos^2\theta}}\\\\=\cos\theta\sin\theta$

LHS=RHS

Hence, proved

#Learn more

Prove this trigonometric identities

brainly.in/question/2708138

Answer 2

Answer:

Consider the provided expression.

(\csc\theta-\sin \theta)(\sec \theta - \cos \theta)=\dfrac{1}{\tan \theta + \cot\theta}(cscθ−sinθ)(secθ−cosθ)=tanθ+cotθ1

Consider the LHS.

\begin{lgathered}=(\csc\theta-\sin \theta)(\sec \theta - \cos \theta)\\=(\frac{1}{\sin\theta}-\sin \theta)(\frac{1}{\cos\theta} - \cos \theta)\\=(\frac{1-\sin^2\theta}{\sin\theta})(\frac{1- \cos^2 \theta}{\cos\theta})\\=(\frac{\cos^2\theta}{\sin\theta})(\frac{\sin^2 \theta}{\cos\theta})\\=\cos\theta\sin\theta\end{lgathered}=(cscθ−sinθ)(secθ−cosθ)=(sinθ1−sinθ)(cosθ1−cosθ)=(sinθ1−sin2θ)(cosθ1−cos2θ)=(sinθcos2θ)(cosθsin2θ)=cosθsinθ

Now Consider the RHS

\begin{lgathered}=\dfrac{1}{\tan \theta + \cot\theta}\\=\dfrac{1}{\frac{\sin\theta}{\cos\theta} +\frac{\cos\theta}{\sin\theta}}\\\\=\dfrac{\cos\theta\sin\theta}{{\sin^2\theta}+{\cos^2\theta}}\\\\=\cos\theta\sin\theta\end{lgathered}=tanθ+cotθ1=cosθsinθ+sinθcosθ1=sin2θ+cos2θcosθsinθ=cosθsinθ

LHS=RHS

Hence, proved