lf asinθ+bcosθ=c, then b+atanθa−btanθ=

asinθ+bcosθ=cacosθ−bsinθ=kSquaring on both sides and adding givesa2+b2=c2+k2k=±a2+b2−c2b+atanθa−btanθ=asinθ+bcosθacosθ−bsinθ=ck=c±a2+b2−c2

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lf a + b = c, then {a - b }{b + a } =

Question

lf $a sin θ + b cos θ = c$ , then $\frac{a - b tan θ}{b + a tan θ} =$

\frac{a ^{2} + b ^{2} - c ^{2}}{c}

- \frac{a ^{2} + b ^{2} - c ^{2}}{c}

\pm \frac{a ^{2} + b ^{2} - c ^{2}}{c}

\pm \frac{a ^{2} + b ^{2} + c ^{2}}{c}

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Alexander Discussed

a sin θ + b cos θ = c

, then

\frac{a - b tan θ}{b + a tan θ} =

8 mins ago

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8 mins ago

a sin θ + b cos θ = c

a cos θ - b sin θ = k

Squaring on both sides and adding gives

a^{2} + b^{2} = c^{2} + k^{2}

k = \pm a^{2} + b^{2} - c^{2}

\frac{a - b tan θ}{b + a tan θ} = \frac{a cos θ - b sin θ}{a sin θ + b cos θ} = \frac{k}{c} = \frac{\pm a ^{2} + b ^{2} - c ^{2}}{c}

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Question Text	lf $a sin θ + b cos θ = c$ , then $\frac{a - b tan θ}{b + a tan θ} =$
Answer Type	Text solution:1
Upvotes	150