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Answers
Answer:
1/5
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Step-by-step explanation:
The value of \sec A-\tan AsecA−tanA is equal to \dfrac{1}{5}
5
1
.
Step-by-step explanation:
We have,
13\sin A =1213sinA=12
∴ \sin A =\dfrac{12}{13}sinA=
13
12
To find, \sec A-\tan AsecA−tanA = ?
∴ \sin A =\dfrac{12}{13}=\dfrac{p}{h}sinA=
13
12
=
h
p
Where p = perpendicular and h = hypotaneous
∴ Base, b = \sqrt{h^{2}-p^{2}}
h
2
−p
2
= \sqrt{13^{2}-12^{2}}=\sqrt{169-144}=\sqrt{25}
13
2
−12
2
=
169−144
=
25
= 5
∴ sec A=\dfrac{h}{b} =\dfrac{13}{5}secA=
b
h
=
5
13
and
\tan A=\dfrac{p}{b} =\dfrac{12}{5}tanA=
b
p
=
5
12
∴ \sec A-\tan AsecA−tanA = \dfrac{13}{5}-\dfrac{12}{5}
5
13
−
5
12
= \dfrac{13-12}{5}
5
13−12
= \dfrac{1}{5}
5
1
∴ The value of \sec A-\tan AsecA−tanA = \dfrac{1}{5}
5
1
Thus, the value of \sec A-\tan AsecA−tanA is equal to \dfrac{1}{5}
5
1
.
Answer:
1/4
Step-by-step explanation:
hope it helps you
....