Question

(c)If A + B = 90°, then prove that: :
sin A sin B cos A sec B + cos A cos B sin A cosec B = 1.
URGENT​

Answer 1

Step-by-step explanation:

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,

\begin{gathered}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\end{gathered}

TanA TanB+

SinA SecB

TanA CotB

−

Cos

2

A

Sin

2

B

=TanA CotA+

SinA CosecA

TanA TanA

−

Sin

2

B

Sin

2

B

=1+Tan

2

A−1=Tan

2

A