Question

If A = ${115}^{\circ }$ and B = ${110}^{\circ }$, then the value of

$\frac{1+cotA}{cotA}.\frac{1+cotB}{cotB}$ is __ .

Answer 1

Like in the previous question, we are given two not so familiar angles here. But their sum, A + B = 225 is a familiar angle,

whose trigonometric ratios are known. In this question also, we are going to consider the expansion of cot(A+B) or cot (A-B).
We want to find $\frac{1+cotA}{cotA}.\frac{1+cotB}{cotB}$
If we expand the numerator, things will be much cleaner.

(1+CotA) (1+CotB) = CotA+CotB+CotACotB+1

Now it is clear why we will consider the expansion of cot (A+B).
Cot (A+B) = $\frac{cotAcotB-1}{cotA+cotB}$
Cot (${225}^{\circ }$) = cot (180+45) = cot (45) = 1

$\Rightarrow$ cot(A+B) = 1 = $\frac{cotAcotB-1}{cotA+cotB}$

$\Rightarrow$ cotA + cotB = cotAcotB - 1

$\Rightarrow$ Numerator = cotAcotB - 1+ cotAcotB + 1

= 2 cotAcotB

$\Rightarrow$ Expression = $\frac{2cotAcotB}{cotAcotB}$ = 2
We can divide each term with cotA and cotB to get
$\frac{1+cotA}{cotA}.\frac{1+cotB}{cotB}$ = (tanA + 1)(tanB + 1)
Proceeding this way will save one or two steps and you can use the expansion of tan (A+B).

Key steps/concepts: (1) Considering the sum of angles (A+B = 225)

(2) Expanding the numerator and guessing the identity to be used.