Question

If A+B =90°then tanAtanB+tanAcotB÷sinAsecB-sin²b÷cos²A is equal to

Answer 1

Answer:

A + B = 90°       =>   A = 90 - B

So Tan A = Cot (90 - A) = Cot B

So Tan B = Cot (90 - B) = Cot A

SecB = Cosec (90 -B) = Cosec A

CosA = Sin (90 -A) = Sin B

substitute these in the LHS,

$tanA \: tanB+ \frac{ tanA \: cotB }{ sinA \: secB} - \frac{ { \sin }^{2} B }{ { \cos }^{2}A } \\ = \tan(A) \cot(B) + \frac{ tanA \: tanB }{ sinA \: cosecB } - \frac{ { \sin }^{2} B }{ { \sin }^{2}A } \\ = 1 + { \tan }^{2} A - 1 \\ = { \tan }^{2} A$