cot A
tan A
+
=sec AcosecA-2 sin Acos A
• Ittan' A 1+cot? A
tan A
cot A
Aac
Answers
Answer: The proof is given below.
Step-by-step explanation: We are given to prove the following trigonometric equality :
\dfrac{\tan A}{1\cot A}+\dfrac{\cot A}{1-\tan A}=1+\sec A\csc A.
1cotA
tanA
+
1−tanA
cotA
=1+secAcscA.
We will be using the following trigonometric formulas :
\begin{gathered}(i)~\tan\theta=\dfrac{\sin\theta}{\cos\theta},\\\\\\(ii)~\sin^2\theta+\cos^2\theta=1.\end{gathered}
(i) tanθ=
cosθ
sinθ
,
(ii) sin
2
θ+cos
2
θ=1.
The proof of the given equality is as follows :
\begin{gathered}L.H.S.\\\\\\\dfrac{\tan A}{1-\cot A}+\dfrac{\cot A}{1-\tan A}\\\\\\=\dfrac{\frac{\sin A}{\cos A}}{1-\frac{\cos A}{\sin A}}+\dfrac{\frac{\cos A}{\sin A}}{1-\frac{\sin A}{\cos A}}\\\\\\=\dfrac{\sin A}{\cos A}\times\dfrac{\sin A}{\sin A-\cos A}+\dfrac{\cos A}{\sin A}\times\dfrac{\cos A}{\cos A-\sin A}\\\\\\=\dfrac{\sin^2 A}{\cos A(\sin A-\cos A)}-\dfrac{\cos^2 A}{\sin A(\sin A-\cos A)}\\\\\\=\dfrac{\sin^3 A-\cos^3 A}{\sin A\cos A(\sin A-\cos A)}\\\\\\=\dfrac{(\sin A-\cos A)(\sin^2 A+\sin A\cos A+\cos^2 A)}{\sin A\cos A(\sin A-\cos A)}\\\\\\=\dfrac{\sin A\cos A+1}{\sin A\cos A}\\\\\\=1+\sec A\csc A\\\\=R.H.S.\end{gathered}
L.H.S.
1−cotA
tanA
+
1−tanA
cotA
=
1−
sinA
cosA
cosA
sinA
+
1−
cosA
sinA
sinA
cosA
=
cosA
sinA
×
sinA−cosA
sinA
+
sinA
cosA
×
cosA−sinA
cosA
=
cosA(sinA−cosA)
sin
2
A
−
sinA(sinA−cosA)
cos
2
A
=
sinAcosA(sinA−cosA)
sin
3
A−cos
3
A
=
sinAcosA(sinA−cosA)
(sinA−cosA)(sin
2
A+sinAcosA+cos
2
A)
=
sinAcosA
sinAcosA+1
=1+secAcscA
=R.H.S.