Question

if tanA=4/3show that under root 1+cosAby 1-cosA=2​

Answer 1

Answer:

Question

if tanA=4/3show that under root 1+cosAby 1-cosA=2

Solution

tan (a) = 4

3

where 4 = perpendicular = p

and 3 = base = b

we have to find hypotenuse (H) first

H^2 = P^2 + B^2

H = √( 16 + 9 )

H = 5

now

cos A = Base

H

= 3/5

now

root 1+cosAby 1-cosA= root 1-3/ 5 divide 1+3/5

we get √ 4

we get 2

hence proved

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Answer 2

Step-by-step explanation:

We are given,

▪tanA = $\frac {4}{3}$

We have to show,

$\frac{ {\sqrt1+cosA}}{1-cosA}$

▪ Oposite side = 4

▪ Adjacent side = 3

∴ Hypotenuse = $\sqrt{{4}^{2}+{3}^{2}}$

= $\sqrt {16+9}$

= $\sqrt{25}$

= $5$

$∴cosA = \frac{3}{5} \\ \\ L.H.S = \frac{ \sqrt{1} + cosA}{1 - cosA} \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: = \frac{ {1 + \frac{3}{5} } }{1 - \frac{3}{5} } \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: = \frac{ { \frac{5 + 3}{5} } }{ \frac{5 - 3}{5} } \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: = \frac{ { \frac{8}{5} } }{ \frac{2}{5} } \\ \: \: \: \: \: \: \: \: \: \: \: \: \: = \frac{8}{5} \times \frac{5}{2} \\ \: \: \: \: \: \: \: \: \: \: \: \: \: = 4$