# Find the value of sin (a+b) if tan a=4/3 and cot b= 5/12, 0^degrees<a<90^degrees and 0^degrees<b<90^degrees ?

May 1, 2018

$\sin \left(a + b\right) = \frac{56}{65}$

#### Explanation:

Given, $\tan a = \frac{4}{3} \mathmr{and} \cot b = \frac{5}{12}$

$\rightarrow \cot a = \frac{3}{4}$

$\rightarrow \sin a = \frac{1}{\csc} a = \frac{1}{\sqrt{1 + {\cot}^{2} a}} = \frac{1}{\sqrt{1 + {\left(\frac{3}{4}\right)}^{2}}} = \frac{4}{5}$

$\rightarrow \cos a = \sqrt{1 - {\sin}^{2} a} = \sqrt{1 - {\left(\frac{4}{5}\right)}^{2}} = \frac{3}{5}$

$\rightarrow \cot b = \frac{5}{12}$

$\rightarrow \sin b = \frac{1}{\csc} b = \frac{1}{\sqrt{1 + {\cot}^{2} b}} = \frac{1}{\sqrt{1 + {\left(\frac{5}{12}\right)}^{2}}} = \frac{12}{13}$

$\rightarrow \cos b = \sqrt{1 - {\sin}^{2} b} = \sqrt{1 - {\left(\frac{12}{13}\right)}^{2}} = \frac{5}{13}$

Now, $\sin \left(a + b\right) = \sin a \cdot \cos b + \cos a \cdot \sin b$

$= \left(\frac{4}{5}\right) \left(\frac{5}{13}\right) + \left(\frac{3}{5}\right) \cdot \left(\frac{12}{13}\right) = \frac{56}{65}$

May 1, 2018

$\sin \left(a + b\right) = \frac{56}{65}$

#### Explanation:

Here,

0^circ < color(violet)( a ) < 90^circ=>I^(st)Quadrant=>color(blue)(All, fns. > 0.

0^circ < color(violet)(b) < 90^circ=>I^(st)Quadrant=>color(blue)(All,fns. > 0

So,

0^circ < color(violet)(a+b) < 180^circ=>I^(st) and II^(nd) Quadrant

=>color(blue)(sin(a+b) > 0

Now,

$\tan a = \frac{4}{3} \implies \sec a = + \sqrt{1 + {\tan}^{2} a} = \sqrt{1 + \frac{16}{9}} = \frac{5}{3}$

:.color(red)(cosa)=1/seca=color(red)(3/5

=>color(red)(sina)=+sqrt(1-cos^2a)=sqrt(1-9/25)=color(red)(4/5

Also,

$\cot b = \frac{5}{12} \implies \csc b = + \sqrt{1 + {\cot}^{2} b} = \sqrt{1 + \frac{25}{144}} = \frac{13}{12}$

:.color(red)(sinb)=1/cscb=color(red)(12/13

=>color(red)(cosb)=+sqrt(1-sin^2b)=sqrt(1-144/169)=color(red)(5/13

Hence,

$\sin \left(a + b\right) = \sin a \cos b + \cos a \sin b$

$\implies \sin \left(a + b\right) = \frac{4}{5} \times \frac{5}{13} + \frac{3}{5} \times \frac{12}{13}$

$\sin \left(a + b\right) = \frac{20}{65} + \frac{36}{65} = \frac{56}{65}$