Question

Prove :tan(45°-A)+cot(45°+A)=(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)

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

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

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

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

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

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

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

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

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

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