Question

if tanx =b/a , then what is the value of acos2x+bsin2x ?

Answer 1

Answer:

refer the above attached pic. Hope it helps u

Answer 2

Answer:

$\red { Value \: of \: acos 2x + b sin2x }\green {=a}$

Step-by-step explanation:

$Given \: tan x = \frac{b}{a}\: ---(1)$

$\red {Value \: of \: acos 2x + b sin2x }$

$= a \left ( \frac{1-tan^{2}x}{1+tan^{2}x}\right) + b \left( \frac{2tanx}{1+tan^{2}x}\right)$

$\boxed { \pink {cos 2x = \frac{1-tan^{2}x}{1+tan^{2}x}}}$

$\boxed { \blue {sin 2x = \frac{2tanx}{1+tan^{2}x}}}$

$= \frac{1}{1+tan^{2}x}\left(a\big(1-tan^{2}x\big)+2btanx\right)$

$= \frac{1}{1+\left(\frac{b}{a}\right)^{2}} \left(a\big(1-\left(\frac{b}{a}\right)^{2}\big)+2b\times \frac{b}{a}\right)$

$= \frac{1}{\frac{a^{2}+b^{2}}{a^{2}}}\left(a\times \frac{a^{2}-b^{2}}{a^{2}}+\frac{2b^{2}}{a}\right)$

$= \frac{a^{2}}{a^{2}+b^{2}}\left(\frac{a^{2}-b^{2}}{a}+\frac{2b^{2}}{a}\right)$

$= \frac{a^{2}}{a^{2}+b^{2}}\left( \frac{a^{2}-b^{2}+2b^{2}}{a}\right)$

$= \frac{a^{2}}{a^{2}+b^{2}}\left( \frac{a^{2}+b^{2}}{a}\right)$

$= \green {a}$

Therefore.,

$\red { Value \: of \: acos 2x + b sin2x }\green {=a}$

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