Prove that:
cos
sin e
+
1-tan 1-coto
sin 0 + cos 0
Answers
Answer:
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
\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+cotθ
1
=
cosθ
sinθ
+
sinθ
cosθ
1
=
sin
2
θ+cos
2
θ
cosθsinθ
=cosθsinθ
LHS=RHS
Hence, proved
Answer:
Step-by-step explanation:
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