please help me in this trig problem.
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please help me in this trig problem.
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Answer:
your solution is as follows
pls mark it as brainliest
Step-by-step explanation:
[tex]to \: find : \\ \frac{a.sin \:θ + b.cos \: θ}{a.sin \:θ - b.cos \: θ } \\ \\ here \: it \: is \: given \: that \\ tan \: θ = \frac{a}{b} \\ \\ so \: let \: here \\ \frac{a.sin \: θ + b.cos \:θ }{a.sin\:θ - b.cos \:θ } = k \\ \\ applying \: componendo \: dividendo \\ on \: both \: sides \\ we \: get \\ \\ \frac{a.sin \: θ + a.sin \:θ }{b.cos \:θ + b.cos \: θ } = \frac{k + 1}{k - 1} \\ \\ \frac{2.a \: sin \:θ }{2.b \: cos \: θ} = \frac{k + 1}{k - 1} \\ \\ = \frac{a}{b} \times tan \: θ \\ \\ = \frac{a}{b} \times \frac{a}{b} \\ \\ = \frac{a {}^{2} }{b {}^{2} } = \frac{k + 1}{k - 1} \\ \\ \: applying \: componendo \: dividendo \\ on \: both \: sides \\ we \: get \\ \\ \frac{a {}^{2} + b {}^{2} }{a {}^{2} - b {}^{2} } = \frac{k + 1 + k - 1}{k + 1 - (k - 1)} \\ \\ = \frac{k + k}{1 + 1} \\ \\ = \frac{2k}{2} \\ \\ = k \\ \\ hence \: then \\ \\ \frac{a.sin \:θ + b.cos \: θ }{a.sin \: θ - b.cos \: θ} = \frac{a {}^{2} + b {}^{2} }{a {}^{2} - b {}^{2} } [/tex]
your solution is as follows
pls mark it as brainliest
Step-by-step explanation:
\begin{gathered}to \: find : \\ \frac{a.sin \:θ + b.cos \: θ}{a.sin \:θ - b.cos \: θ } \\ \\ here \: it \: is \: given \: that \\ tan \: θ = \frac{a}{b} \\ \\ so \: let \: here \\ \frac{a.sin \: θ + b.cos \:θ }{a.sin\:θ - b.cos \:θ } = k \\ \\ applying \: componendo \: dividendo \\ on \: both \: sides \\ we \: get \\ \\ \frac{a.sin \: θ + a.sin \:θ }{b.cos \:θ + b.cos \: θ } = \frac{k + 1}{k - 1} \\ \\ \frac{2.a \: sin \:θ }{2.b \: cos \: θ} = \frac{k + 1}{k - 1} \\ \\ = \frac{a}{b} \times tan \: θ \\ \\ = \frac{a}{b} \times \frac{a}{b} \\ \\ = \frac{a {}^{2} }{b {}^{2} } = \frac{k + 1}{k - 1} \\ \\ \: applying \: componendo \: dividendo \\ on \: both \: sides \\ we \: get \\ \\ \frac{a {}^{2} + b {}^{2} }{a {}^{2} - b {}^{2} } = \frac{k + 1 + k - 1}{k + 1 - (k - 1)} \\ \\ = \frac{k + k}{1 + 1} \\ \\ = \frac{2k}{2} \\ \\ = k \\ \\ hence \: then \\ \\ \frac{a.sin \:θ + b.cos \: θ }{a.sin \: θ - b.cos \: θ} = \frac{a {}^{2} + b {}^{2} }{a {}^{2} - b {}^{2} } \end{gathered}
tofind:
a.sinθ−b.cosθ
a.sinθ+b.cosθ
hereitisgiventhat
tanθ=
b
a
solethere
a.sinθ−b.cosθ
a.sinθ+b.cosθ
=k
applyingcomponendodividendo
onbothsides
weget
b.cosθ+b.cosθ
a.sinθ+a.sinθ
=
k−1
k+1
2.bcosθ
2.asinθ
=
k−1
k+1
=
b
a
×tanθ
=
b
a
×
b
a
=
b
2
a
2
=
k−1
k+1
applyingcomponendodividendo
onbothsides
weget
a
2
−b
2
a
2
+b
2
=
k+1−(k−1)
k+1+k−1
=
1+1
k+k
=
2
2k
=k
hencethen
a.sinθ−b.cosθ
a.sinθ+b.cosθ
=
a
2
−b
2
a
2
+b
2