If $sin(theta)+cos(theta)=(sqrt2) sin(90^@-theta)$ ,find $tan(theta)$ ?

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Answer 1 · 2017-08-16T16:59:19

$tan(theta)=sqrt2 -1$

Given: $sin(theta)+cos(theta)=(sqrt2) sin(90^@-theta)$

Use the identitity $sin(A-B) = sin(A)cos(B)-cos(A)sin(B)$ where $A = 90^@$ and $B = theta$

$sin(theta)+cos(theta)=(sqrt2)(sin(90^@)cos(theta)-cos(90^@)sin(theta))$

Use the facts that $sin(90^@) = 1 and cos(90^@)=0$

$sin(theta)+cos(theta)=(sqrt2)cos(theta)$

Divide both sides of the equation by $cos(theta)$ :

$sin(theta)/cos(theta)+ 1=sqrt2$

Use the identity $sin(theta)/cos(theta) = tan(theta)$

$tan(theta)+ 1=sqrt2$

Subtract 1 from both sides:

$tan(theta)=sqrt2 -1$

If sin(theta)+cos(theta)=(sqrt2) sin(90^@-theta)sin(theta)+cos(theta)=(sqrt2) sin(90^@-theta),find tan(theta)tan(theta)?