Question

in triangle abc right angled at b where ab = 8 bc = 6 find: tanA + cot A - cos A​

Answer 1

Given,

AB = 8

BC = 6

Now, From Pythagoras Theorum :-

$\implies AC = \sqrt{{AB}^{2}+{BC}^{2}}$

$\implies AC = \sqrt{{8}^{2}+{6}^{2}}$

$\implies AC = \sqrt{100}$

$\implies AC = 10$

Again for Expression,

With respect to <A

=> p = BC = 6

=> b = AB = 8

=> h = AC = 10

=> Tan A + Cot A - Cos A

$\implies \dfrac{p}{b}+\dfrac{b}{p}-\dfrac{b}{h}$

$\implies \dfrac{6}{8}+\dfrac{8}{6}-\dfrac{8}{10}$

$\implies \dfrac{90+160-96}{120}$

$\Implies \dfrac{154}{120}$

$\implies \dfrac{77}{60}$