Question

plz. solve this it is very important​

Answer 1

Answer:

Step-by-step explanation:

Answers

duragpalsingh

Duragpalsingh Maths AryaBhatta

Hey there! ☺☻☺

Given,

4 tan = 3

∴ tan = 3/4

As we know,

Tan = Perpendicular / base

Tan = 3/4

Now,

Hypotenuse = √3² + 4² = √9+16 = √25 = 5

Using this, we can find:

Sin = Perpendicular / Hypotenuse = 3/5

Cos = Base / Hypotenuse = 4/5

ATQ,

(4sin - cos + 1)/(4sin + cos - 1)

= (4 × 3/5 - 4/5 + 1)/(4 × 3/5 + 4/5 - 1) [putting values]

= (12 - 4 + 5)/(12 + 4 - 5)

= 13/11

Answer 2

[ 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

Hope this helps u