Math, asked by nagunagendra858, 8 months ago

Prove that:
cos
sin e
+
1-tan 1-coto
sin 0 + cos 0​

Answers

Answered by kariabhumi02
0

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

Answered by singlahunar1809
1

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

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