Question

cos e
(ii)
sin e
1-tan 0 1 - coto
+
= cos 0 + sin 0.​

Answer 1

Consider the provided expression.

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

tanθ+cotθ

1

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)

=(

sinθ

1

−sinθ)(

cosθ

1

−cosθ)

=(

sinθ

1−sin

2

θ

)(

cosθ

1−cos

2

θ

)

=(

sinθ

cos

2

θ

)(

cosθ

sin

2

θ

)

=cosθsinθ

Now Consider the RHS

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