- ) sW‘::': ) SHE[EAD °G ० 0 एप हा

1.	- ) sW‘::': ) SHE[EAD °G ० 0 एप हा
Answer» Given,4 tan = 3∴ tan = 3/4 As we know,Tan= Perpendicular / baseTan= 3/4Now,Hypotenuse =√3² + 4² =√9+16 =√25 = 5 Using this, we can find:Sin = Perpendicular / Hypotenuse = 3/5Cos = 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»

Given,4 tan = 3∴ tan = 3/4

As we know,Tan= Perpendicular / baseTan= 3/4Now,Hypotenuse =√3² + 4² =√9+16 =√25 = 5

Using this, we can find:Sin = Perpendicular / Hypotenuse = 3/5Cos = 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

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