21.
Prove the following identity:
sinA/secA+tanA-1 + cosA/cosecA+cotA-1 = 1
Answers
Step-by-step explanation:
The function is
\dfrac{\sin A}{\sec A+\tan A-1}+\dfrac{\cos A}{cosec A+cot A-1}=1
secA+tanA−1
sinA
+
cosecA+cotA−1
cosA
=1
We need to prove the left hand side equal right hand side
Using left hand side
=\dfrac{\sin A}{\sec A+\tan A-1}+\dfrac{\cos A}{cosec A+cot A-1}=
secA+tanA−1
sinA
+
cosecA+cotA−1
cosA
=\dfrac{\sin A}{\dfrac{1}{\cos A}+\dfrac{\sin A}{\cos A}-1}+\dfrac{\cos A}{\dfrac{1}{\sin A}+\dfrac{\cos A}{\sin A}-1}=
cosA
1
+
cosA
sinA
−1
sinA
+
sinA
1
+
sinA
cosA
−1
cosA
=\dfrac{\sin A}{\dfrac{1+\sin A-\cos A}{\cos A}}+\dfrac{\cos A}{\dfrac{1+\cos A-\sin A}{\sin A}}=
cosA
1+sinA−cosA
sinA
+
sinA
1+cosA−sinA
cosA
=\dfrac{\sin A\times\cos A}{1+\sin A-\cos A}+\dfrac{\cos A\times\sin A}{1+\cos A-\sin A}=
1+sinA−cosA
sinA×cosA
+
1+cosA−sinA
cosA×sinA
=\sin A\cos A(\dfrac{1}{1+\sin A-\cos A}+\dfrac{1}{1+\cos A-\sin A})=sinAcosA(
1+sinA−cosA
1
+
1+cosA−sinA
1
)
=\sin A\cos A(\dfrac{(1+\cos A-\sin A)+(1+\sin A-\cos A)}{(1+\cos A-\sin A)(1+\sin A-\cos A)})=sinAcosA(
(1+cosA−sinA)(1+sinA−cosA)
(1+cosA−sinA)+(1+sinA−cosA)
)
=\sin A\cos A(\dfrac{2}{(1+\cos A-\sin A)(1+\sin A-\cos A)})=sinAcosA(
(1+cosA−sinA)(1+sinA−cosA)
2
)
=\sin A\cos A(\dfrac{2}{1+\sin A-\cos A+\cos A+\sin A\cos A-\cos^{2}A-\sin A-\sin^{2}A+\cos A\sin A})=sinAcosA(
1+sinA−cosA+cosA+sinAcosA−cos
2
A−sinA−sin
2
A+cosAsinA
2
)
=\sin A\cos A(\dfrac{2}{1-(\sin^{2}A+\cos^{2})+2\cos A\sin A})=sinAcosA(
1−(sin
2
A+cos
2
)+2cosAsinA
2
)
Here, \sin^{2}A+\cos^{2}A=1sin
2
A+cos
2
A=1
=\sin A\cos A(\dfrac{2}{1-1+2\cos A\sin A})=sinAcosA(
1−1+2cosAsinA
2
)
=\sin A\cos A\times\dfrac{1}{\cos A\sin A}=sinAcosA×
cosAsinA
1
=1=1