Question

If A + B = 90°, then prove that
tan A tan B+tan A cot B sin? B
sin A sec B
cos? A
=tan A.​

Answer 1

Answer:

A + B = 90° => A = 90 - B

A + B = 90° => A = 90 - BSo Tan A = Cot (90 - A) = Cot B

A + B = 90° => A = 90 - BSo Tan A = Cot (90 - A) = Cot BSo Tan B = Cot (90 - B) = Cot A

A + B = 90° => A = 90 - BSo Tan A = Cot (90 - A) = Cot BSo Tan B = Cot (90 - B) = Cot ASecB = Cosec (90 -B) = Cosec A

A + B = 90° => A = 90 - BSo Tan A = Cot (90 - A) = Cot BSo Tan B = Cot (90 - B) = Cot ASecB = Cosec (90 -B) = Cosec ACosA = Sin (90 -A) = Sin B

A + B = 90° => A = 90 - BSo Tan A = Cot (90 - A) = Cot BSo Tan B = Cot (90 - B) = Cot ASecB = Cosec (90 -B) = Cosec ACosA = Sin (90 -A) = Sin Bsubstitute these in the LHS,

A + B = 90° => A = 90 - BSo Tan A = Cot (90 - A) = Cot BSo Tan B = Cot (90 - B) = Cot ASecB = Cosec (90 -B) = Cosec ACosA = Sin (90 -A) = Sin Bsubstitute these in the LHS,TanA\ TanB+\frac{TanA\ CotB}{SinA \ SecB}-\frac{Sin^2B}{Cos^2A}\\\\=TanA\ CotA + \frac{TanA\ TanA}{SinA\ CosecA}-\frac{Sin^2B}{Sin^2B}\\\\=1+Tan^2A - 1=Tan^2A

Answer 2

Answer:

Some relations you should know:

If A+B=90°, then:

→ tanB = cotA

→ tanA = cotB

→ sinB = cosA

→ sinA = cosB

$\begin{gathered} \sf{sqrt{ \frac{tanA.tanB-tanA.cotB}{sinA.secB}}- \sf{\frac{sin^2B}{cos^2A} } }\\ \\ = \sqrt{ \frac{tanA.cotA-tanA.tanA}{sinA.(1/cosB)}- \frac{sin^2B}{sin^2B} } \\ \\ =\sqrt{ \frac{tanA.(1/tanA)-tan^2A}{sinA.(1/sinA)}- \frac{sin^2B}{sin^2B} } \\ \\ =\sqrt{ \frac{1-tan^2A}{1}-1 }\\ \\ = \sqrt{-tan^2A} \\ \\=\pm i\ tanA\end{gathered}$

It's not coming tanA.