simplify (1+tanA+secA) (1+cotA-cosecA)
Answers
Step-by-step explanation:
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Answer:
Solution: Given: (1+ tan A + sec A) (1+ cot A - cosec A)
The given expression (1+tanA+secA)(1+cotA-cosecA)(1+tanA+secA)(1+cotA−cosecA)
SOLUTION :
Given expression is (1+tanA+secA)(1+cotA-cosecA)(1+tanA+secA)(1+cotA−cosecA)
Now solving the given expression as below
(1+tanA+secA)(1+cotA-cosecA)(1+tanA+secA)(1+cotA−cosecA)
=(1+\frac{sinA}{cosA}+\frac{1}{cosA})(1+\frac{cosA}{sinA}-\frac{1}{sinA})=(1+
cosA
sinA
+
cosA
1
)(1+
sinA
cosA
−
sinA
1
)
By using the trignometric formulae :
i) tanx=\frac{sinx}{cosx}tanx=
cosx
sinx
ii) secx=\frac{1}{cosx}secx=
cosx
1
iii) cotx=\frac{cosx}{sinx}cotx=
sinx
cosx
iv) cosecx=\frac{1}{secx}cosecx=
secx
1
=(\frac{(cosA+sinA)+1}{cosA})(\frac{(sinA+cosA)-1}{sinA})=(
cosA
(cosA+sinA)+1
)(
sinA
(sinA+cosA)−1
)
=(\frac{(cosA+sinA)+1}{cosA})(\frac{(cosA+sinA)-1}{sinA})=(
cosA
(cosA+sinA)+1
)(
sinA
(cosA+sinA)−1
)
By using the Algebraic identity :
(a+b)(a-b)=a^2-b^2(a+b)(a−b)=a
2
−b
2
=\frac{(cosA+sinA)^2-1^2}{cosA sinA}=
cosAsinA
(cosA+sinA)
2
−1
2
By using the Algebraic identity :
(a+b)^2=a^2+2ab+b^2(a+b)
2
=a
2
+2ab+b
2
=\frac{cos^2A+sin^2A+2cosA sinA-1}{cosA sinA}=
cosAsinA
cos
2
A+sin
2
A+2cosAsinA−1
By using the trignometric identity :
cos^2x+sin^2x=1cos
2
x+sin
2
x=1
=\frac{1+2cosA sinA-1}{cosA sinA}=
cosAsinA
1+2cosAsinA−1
=\frac{2cosA sinA}{cosA sinA}=
cosAsinA
2cosAsinA
= 2
∴ (1+tanA+secA)(1+cotA-cosecA)=2(1+tanA+secA)(1+cotA−cosecA)=2
∴ the evaluated value for the given expression (1+tanA+secA)(1+cotA-cosecA)(1+tanA+secA)(1+cotA−cosecA) is 2