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