Question

Cot (a+45°)-tan (a-45°)=2cos2a/1+sin2a

Answer 1

Answer:

tan(45-A)+cot(45+A)=\frac{2cos2A}{1+sin2A}

Step-by-step explanation:

Formula used:

tan(A-B)=\frac{tanA-tanB}{1+tanA tanB}

tan(A+B)=\frac{tanA+tanB}{1-tanA tanB}

cos2A=cos^2A-sin^2A

sin2A=2\:sinA\:cosA

Now,

tan(45-A)+cot(45+A)

=tan(45-A)+\frac{1}{tan(45+A)}

=\frac{tan45-tanA}{1+tan45\:tanA}+\frac{1}{\frac{tan45+tanA}{1-tan45 tanA}}

=\frac{1-tanA}{1+tanA}+\frac{1}{\frac{1+tanA}{1-tanA}}

=\frac{1-tanA}{1+tanA}+\frac{1-tanA}{1+tanA}

=2\frac{1-tanA}{1+tanA}

Multiply both numerator and denominator by (1+tanA)

=2\frac{1-tanA}{1+tanA}*\frac{1+tanA}{1+tanA}

=2\frac{1^2-tan^2A}{(1+tanA)^2}

=2\frac{1-\frac{sin^2A}{cos^2A}}{(1+\frac{sinA}{cosA})^2}

=2\frac{\frac{cos^2A-sin^2A}{cos^2A}}{(\frac{cosA+sinA}{cosA})^2}

=2\frac{\frac{cos^2A-sin^2A}{cos^2A}}{\frac{(cosA+sinA)^2}{cos^2A}}

=2\frac{cos^2A-sin^2A}{(cosA+sinA)^2}

=2\frac{cos2A}{cos^2A+sin^2A+2\:sinA\:cosA}

=2\frac{cos2A}{1+2\:sinA\:cosA}

=2\frac{cos2A}{1+sin2A}

tan(45-A)+cot(45+A)=\frac{2cos2A}{1+sin2A}

Step-by-step explanation: