Question

27) If tan0 = coto and sin0.seco = 1, then the value of tan²0+ sin 0 + coto
is equal to​

Answer 1

Answer:

(cscθ−sinθ)(secθ−cosθ)=

tanθ+cotθ

1

Proved.

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+cotθ

1

Consider the LHS.

\begin{gathered}=(\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{gathered}

=(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{gathered}=\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{gathered}

=

tanθ+cotθ

1

=

cosθ

sinθ

+

sinθ

cosθ

1

=

sin

2

θ+cos

2

θ

cosθsinθ

=cosθsinθ

LHS=RHS

Hence, proved

#Learn more

Prove this trigonometric identities

brainly.in/question/2708138