Given 3sin- -5cos- -5 then the value of 3cos- -5sin- 2 is

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Trigonometric Identities

Given 3sinβ...

Question

Given $3 sin β + 5 cos β = 5$ then the value of ${(3 cos β - 5 sin β)}^{2}$ is

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\frac{9}{5}

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\frac{1}{3}

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\frac{1}{9}

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Solution

The correct option is A 9
$3 s i n β + 5 c o s β = 5$
Squaring both sides, we have
$9 s i n^{2} β + 25 c o s^{2} β + 30 s i n β c o s β = 25$
Now we have to find the value of
$(3 c o s β - 5 s i n β)^{2}$
$9 c o s^{2} β + 25 s i n^{2} β - 30 s i n β c o s β$ putting the value
$30 s i n β c o s β = 25 - 9 s i n^{2} β - 25 c o s^{2} β$ we have
$9 c o s^{2} β + 25 s i n^{2} β - 30 s i n β c o s β = 9 + 25 - 25 = 9$
using $s i n^{2} β + c o s^{2} β = 1$

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Given 3sinβ+5cosβ=5 then the value of (3cosβ−5sinβ)2 is

Given $3 sin β + 5 cos β = 5$ then the value of ${(3 cos β - 5 sin β)}^{2}$ is