Question

If cos theta =3 by 5 then show. That 5 sin theta -3 tan theta = 0

Answer 1

Given:

• cos Ф = 3/5

To show,

• 5 sinФ - 3 tan Ф = 0

Now,

We know that,

• sinФ = P/H
• cosФ = B/H
• tanФ = P/B
• cotФ = B/P
• secФ = H/B
• cosecФ = H/P

[H = Hypotenuse, P = perpendicular, and B = Base]

So,

• cos Ф = 3/5 = B/H

Let B = 3k, and H = 5k

Now

• Hypotenuse² = Perpendicular² + Base²

⇒ (5k)² = (Perpendicular)² + (3k)²

⇒ (5k)² - (3k)² = (Perpendicular)²

⇒ 25k² - 9k² = (Perpendicular)²

⇒ 16k² = (Perpendicular)²

[Squaring both sides we get]

⇒ 4k = Perpendicular

Now,

• sinФ - 3 tan Ф = 0

LHS

= 5 sinФ - 3 tan Ф

= 5 *(P/H) - 3 * (P/B)

= 5 *(4k/5k) - 3 * (4k/3k)

= 5 * (4/5) - 3 * (4/3)

= 4 - 4

= 0

Hence, proved.

Answer 2

${ \underline{ \underline{ \sf{ \red{Question : }}}}}$

$\tt \bold{If \: \: cos \: \: \theta \: \: = \: \: \frac{3}{5} \: \: then \: \: show \: \: that \: \: 5 \: \: sin \: \: \ theta \: \: -3 \: \: tan \: \: \theta =0}$

${ \underline{ \underline{ \sf{ \red{Solution : }}}}}$

Given:

$\tt \bold \: cos \: \: \theta \: = \frac{3}{5} = \frac{Adjecent \: \: side }{hypotenuse \: \: side}$

• Adjecent side = 3
• Hypotenuse side= 5

Now ,

$\tt \bold \: opposite \: \: side \: \: = \: \: \sqrt{ {(hypotenuse \: \: side)}^{2} - {(adjecent \: side)}^{2} } \\ \\ \tt \bold \: opposite \: \: side \: \: = \: \: \sqrt{ {(5)}^{2} - {(3)}^{2} } \\ \\ \tt \bold \: opposite \: \: side \: \: = \: \: \sqrt{25 - 9} \\ \\ \tt \bold \: opposite \: \: side \: \: = \: \: \sqrt{16} \\ \\ \tt \bold \: opposite \: \: side \: \: = \: \:4$

And,

$\implies \tt \bold \: sin \: \theta \: = \frac{opposite \: \: side}{hypotenuse \: \: side} = \frac{4}{5} \\ \\ \implies \tt \bold \: tan \: \theta \: = \frac{opposite \: \: side}{adjecent\: \: side} = \frac{4}{3}$

Then,

$\implies \sf \large \: 5 \: \: sin \: \theta \: \: - 3 \: \: tan \: \theta \\ \\ \implies \sf \large \cancel5 \left( \frac{4}{ \cancel5} \right) - \cancel3 \: \left( \frac{4}{ \cancel3} \right) \\ \\ \implies \sf \large4 - 4 \\ \\ \implies \sf \large0$

LHS = RHS

Hence proved...