Question

if 4 tan theta=3 evaluate (4sin theta -cos theta +1 appon 4sin theta + cos theta -1

Answer 1

↑ Here is your answer ↓
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Let ∠A be '∅' (see figure.)

$4tan∅ = 3$

$tan∅ = \frac{3}{4}$

$\frac{Opposite}{Adjacent} = \frac{3}{4}$

So, hypothenuse will be 5 by Pythagoras theorem.

Now,

$\frac{4sin∅ - cos∅ + 1}{4sin∅ + cos∅ - 1}$

$\frac{ 4 * \frac{3}{5} - \frac{4}{5} + 1 }{ 4 * \frac{3}{5} + \frac{4}{5} - 1 }$

$\frac{ \frac{12- 4 + 5}{5} }{ \frac{12 +4 - 5}{5}}$

5 will cancel out,

$= \frac{13}{11}$

$= 1.18$

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Glad to help you.
@EmadAhamed

Answer 2

Solution :

[ Here I am using A instead of

theta ]

It is given that ,

4tanA = 3 ---( 1 )

tanA = 3/4

=> tan² A = 9/16

=> sec²A - 1 = 9/16

=> sec²A = 9/16 + 1

=> sec²A = ( 9 + 16)/16

=> Sec²A = 25/16

=> SecA = 5/4 ---( 2 )

Now ,

(4sinA-cosA+1)/(4sinA+cosA-1)

Divide numerator and denominator

With cosA, we get

(4tanA-1+secA)/(4tanA+1-secA)

= (3-1+5/4)/(3+1-5/4) [from(1)&(2)]

= (2+5/4)/(4-5/4)

Divide numerator and denominator

With 4 , we get

= ( 8 + 5 )/( 16 - 5 )

= 13/11

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