If cos-cos-sin-sin-1- then the value of cos 3-cos-sin 3-sin- is

Cosαcosβ+sinαsinβ= 1 Let cosαcosβ = a, then nbsp; sinαsinβ= (1+a) Now, cos^3 βcosα+sin^3 βsinα = cos^2 β nbsp;a sin^2β(1+a) nbsp;= cos^2 β nbsp;a 1 ...

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Standard XII

Mathematics

Basic Trigonometric Identities

If cosα/cosβ+...

Question

If $\frac{cos α}{cos β} + \frac{sin α}{sin β} = - 1,$ then the value of $\frac{{cos}^{3} β}{cos α} + \frac{{sin}^{3} β}{sin α}$ is

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Solution

$\frac{cos α}{cos β} + \frac{sin α}{sin β} = - 1$

Let $\frac{cos α}{cos β} = a$ , then
$\frac{sin α}{sin β} = - (1 + a)$

Now,
$\frac{{cos}^{3} β}{cos α} + \frac{{sin}^{3} β}{sin α} = \frac{{cos}^{2} β}{a} - \frac{{sin}^{2} β}{(1 + a)} = \frac{{cos}^{2} β}{a} - \frac{1 - {cos}^{2} β}{(1 + a)} = {cos}^{2} β [\frac{1}{a} + \frac{1}{1 + a}] - \frac{1}{1 + a} = {cos}^{2} β [\frac{1 + 2 a}{a (1 + a)}] - \frac{1}{1 + a} \dots (1)$

We know that,
${cos}^{2} α + {sin}^{2} α = 1 \Rightarrow a^{2} {cos}^{2} β + (1 + a)^{2} {sin}^{2} β = 1 \Rightarrow a^{2} {cos}^{2} β + (1 + a^{2} + 2 a) (1 - {cos}^{2} β) = 1 \Rightarrow - (1 + 2 a) {cos}^{2} β = 1 - (1 + a^{2} + 2 a) \Rightarrow {cos}^{2} β = \frac{a^{2} + 2 a}{1 + 2 a} \Rightarrow {cos}^{2} β (1 + 2 a) = a^{2} + 2 a \Rightarrow \frac{{cos}^{2} β (1 + 2 a)}{a (a + 1)} - \frac{1}{1 + a} = \frac{a + 2}{a + 1} - \frac{1}{1 + a} \Rightarrow \frac{{cos}^{2} β (1 + 2 a)}{a (a + 1)} - \frac{1}{1 + a} = 1$

From equation $(1)$ , we get
$\frac{{cos}^{3} β}{cos α} + \frac{{sin}^{3} β}{sin α} = 1$

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If cosαcosβ+sinαsinβ=−1, then the value of cos3βcosα+sin3βsinα is

If $\frac{cos α}{cos β} + \frac{sin α}{sin β} = - 1,$ then the value of $\frac{{cos}^{3} β}{cos α} + \frac{{sin}^{3} β}{sin α}$ is