а
If tan =
then find
cos + sine
cos - sino
b ь
Answers
Answer:
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Answer:
The result of \frac { \cos \theta + \sin \theta } { \cos \theta - \sin \theta }
cosθ−sinθ
cosθ+sinθ
is found out to be \frac { \mathrm { a } + \mathrm { b } } { \mathrm { b } - \mathrm { a } }
b−a
a+b
To find:
The result of \frac { \cos \theta + \sin \theta } { \cos \theta - \sin \theta }
cosθ−sinθ
cosθ+sinθ
when \tan \theta = \frac { \mathrm { a } } { \mathrm { b } }tanθ=
b
a
Solution:
Given that the value of \tan \theta = \frac { \mathrm { a } } { \mathrm { b } }tanθ=
b
a
The value of \frac { \cos \theta + \sin \theta } { \cos \theta - \sin \theta }
cosθ−sinθ
cosθ+sinθ
is determined by dividing the numerator and denominator by \cos \thetacosθ
\begin{gathered}\begin{array} { c } { \frac { \cos \theta + \sin \theta } { \cos \theta - \sin \theta } = \frac { \frac { \cos \theta } { \cos \theta } + \frac { \sin \theta } { \cos \theta } } { \frac { \cos \theta } { \cos \theta } - \frac { \sin \theta } { \cos \theta } } } \\\\ { = \frac { 1 + \tan \theta } { 1 - \tan \theta } } \end{array}\end{gathered}
cosθ−sinθ
cosθ+sinθ
=
cosθ
cosθ
−
cosθ
sinθ
cosθ
cosθ
+
cosθ
sinθ
=
1−tanθ
1+tanθ
We know that \tan \theta = \frac { \mathrm { a } } { \mathrm { b } }tanθ=
b
a
, as per the given question,
\begin{gathered}\begin{aligned} & = \frac { 1 + \frac { \mathrm { a } } { \mathrm { b } } } { 1 - \frac { \mathrm { a } } { \mathrm { a } } } \\\\ & = \frac { \frac { \mathrm { b } + \mathrm { a } } { \mathrm { b } } } { \frac { \mathrm { b } - \mathrm { a } } { \mathrm { b } } } \\\\ = & \frac { \mathrm { b } + \mathrm { a } } { \mathrm { b } - \mathrm { a } } \\ \frac { \cos \theta + \sin \theta } { \cos \theta - \sin \theta } & = \frac { \mathrm { a } + \mathrm { b } } { \mathrm { b } - \mathrm { a } } \end{aligned}\end{gathered}
=
cosθ−sinθ
cosθ+sinθ
=
1−
a
a
1+
b
a
=
b
b−a
b
b+a
b−a
b+a
=
b−a
a+b