Question

If secQ=5/4, Find the value of
SinQ-2CosQ/TanQ-CotQ

Answer 1

Given :-

• secQ = (5/4)

To Find :-

• (SinQ - 2CosQ) /(TanQ - CotQ)

Formula used :-

• sin theta = Perpendicular/Hypotenuse
• cos theta = Base/Hypotenuse
• tan theta = Perpendicular/Base
• cosec theta = Hypotenuse/Perpendicular
• sec theta = Hypotenuse/Base
• cot theta = Base/Perpendicular

Solution :-

→ secQ = (5/4) = Hypotenuse/Base

So,

→ Hypotenuse = 5

→ Base = 4

So, By Pythagoras Theoram ,

→ Perpendicular = √(5)² - (4)²

→ Perpendicular = √(25 - 16)

→ Perpendicular = √9

→ Perpendicular = 3 .

_______________

So,

→ SinQ = Perpendicular/Hypotenuse = 3/5

→ cosQ = Base/Hypotenuse = 4/5

→ TanQ = Perpendicular/Base = 3/4

→ CotQ = Base/Perpendicular = 4/3

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Putting All Values now, we get :-

→ (SinQ - 2CosQ) /(TanQ - CotQ)

→ [ (3/5) - 2*(4/5) ] / [ (3/4) - (4/3) ]

→ [ (3- 8)/(5) ] [ ( 9 - 16) / (12) ]

→ (-5/5) / (-7/12)

→ (-1) * (-12/7)

→ (12/7) (Ans).

Answer 2

Answer:

given:

secQ=5/4

we know that secQ=H/B

by Pythagoras theoram

=>(Hypotenuse)²=(Base)²+(perpendicular)²

=>(5)²=(4)²+(p)²

=>25=16+(p)²

=>25-16=(p)²

=>9=(p)²

=>p=√9

=>p=3

Now,

SinQ=p/h =3/5

CosQ=b/h=4/5

TanQ=p/b=3/4

CotQ=b/p=4/3

Now put the value of these in the given equation,

$\implies\frac{sinq - 2cosq}{tanq - cotq} \\ \\ \implies \: \frac{ \frac{3}{5 } - 2 \times \frac{4}{5} }{ \frac{3}{4} - \frac{4}{3} } \\ \\ \implies \: \frac{ \frac{3}{5} - \frac{8}{5} }{ \frac{9 - 16}{12} } \\ \\ \implies \frac{ - 5}{5} \times \frac{12}{ - 7} \\ \\ \implies \: \frac{12}{7}$