Prove that SinA (1+ TanA) + Cos A (1+ CotA) =SecA + Cosec A
Answers
Answered by
1
Answer:
sinA(1+tanA)+cosA(1+cotA)
=sinA+sinAtanA+cosA+cosAcotA
=sinA+sinA
cosA
sinA
+cosA+cosA
sinA
cosA
[∵tanA=
cosA
sinA
,cotA=
sinA
cosA
]
=sinA+
cosA
sin
2
A
+cosA+
sinA
cos
2
A
=
sinAcosA
sin
2
Acos+sin
3
A+cos
2
AsinA+cos
3
A
=
sinAcosA
sinAcosA(sinA+cosA)sin
3
A+cos
3
A
=
sinAcosA
sinAcosA(sinA+cosA)(sin
2
A+cos
2
A−cosAsinA)
[∵a
3
+b
3−(a+b)(a
2
−ab+b
2
)
]
=
sinacosA
(sinA+cosA)sinAcosA+sin
2
A+cos
2
A−sina+cosA
=
sinacosA
(sinA+cosA).1
[∵sin
2
a+cos
2
A=1
]
=
sinAcosA
sinA
+
sinAcosA
cosA
=
CosA
1
+
sina
1
=secA+cosec A.
Step-by-step explanation:
Answered by
5
Answer:
secA+cosecA=secA+cosecA
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