Question

if cotθ = a/b, then the value of ,(a sin θ - b cos θ)/(a sin θ + b cos θ) is? ​

Answer 1

Answer:-

Given:

cot θ = a/b

We know,

• cot θ = cos θ / sin θ

So,

⟶ cos θ / sin θ = a/b

⟶ b cos θ = a sin θ -- equation (1)

We have to find the value of ,

(a sin θ - b cos θ) / (a sin θ + b cos θ)

Writing a sin θ as b cos θ from equation (1) we get,

⟶ (b cos θ - b cos θ) / (b cos θ + b cos θ)

⟶ 0/2b cos θ

⟶ 0

∴ The value of (a sin θ - b cos θ) / (a sin θ + b cos θ) is "0" (zero).

Additional Information:-

• sin² θ + cos² θ = 1

• sec² θ - tan² θ = 1

• cosec² θ - cot² θ = 1

• sin A = 1/Cosec A

• cos A = 1/sec A

• tan A = 1/cot A

• tan A = sin A/cos A

• cot A = 1/tan A

• cot A = cos A/sin A

Answer 2

${ \huge{ \boxed{ \overline{ \mid{ \sf{Solution:-}}}}}}$

$\sf{We \: know,} \\ \\ </p><p></p><p> \sf{cot \theta = cos \theta/ sin \theta}$

$\sf{so}$

$\sf{Writing \: a \: sin \: as \: b \: cos \: from \: equation \: (1) \: we \: get,} \\ \\ </p><p></p><p> \sf{- ( \frac{b \: cos \theta - b \: cos \theta}{b \: cos \theta + b \: cos \theta} )} \\ \\ </p><p></p><p> \sf{ \frac{0}{2 \: b \: cos \theta} }$

$\\ \sf{ \frac{ - cos \theta}{sin \theta} = \frac{a}{b} } \\ \\ </p><p></p><p> \sf{- cos \theta = a \: \: \: \: (sin 0 -- equation (1))}$

$\sf{We \: have \: to \: find \: the \: value \: of,} \\ \\ </p><p></p><p> \sf{( \frac{a \: sin \theta - b \: cos \theta}{a \: sin \theta + b \: cos \theta} )}$

$\implies \huge \sf {\boxed {\boxed{ = 0}}}$

$\sf \underline{the \: value \: is \: {\boxed {\boxed 0}}}$

${ \huge{ \boxed{ \overline{ \mid { \sf{Extra \: info:-}}}}}}$

$\boxed{\begin{minipage}{6cm} Important Trigonometric identities :- \\ \\ \: \: 1)\:\sin^2\theta+\cos^2\theta=1 \\ \\ 2)\:\sin^2\theta= 1-\cos^2\theta \\ \\ 3)\:\cos^2\theta=1-\sin^2\theta \\ \\ 4)\:1+\cot^2\theta=\text{cosec}^2 \, \theta \\ \\5)\: \text{cosec}^2 \, \theta-\cot^2\theta =1 \\ \\ 6)\:\text{cosec}^2 \, \theta= 1+\cot^2\theta \\\ \\ 7)\:\sec^2\theta=1+\tan^2\theta \\ \\ 8)\:\sec^2\theta-\tan^2\theta=1 \\ \\ 9)\:\tan^2\theta=\sec^2\theta-1 \end{minipage}}$