If Tan A =3/4 find the value of secA and CosA
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tan A=3/4.
Sec^2 A= 1+tan^2 A= 1+(3/4)^2=1+9/16=25/16.
Sec A=(25/16)^0.5= +/-5/4.
Cot A=1/tan A=1/(3/4)=4/3.
cosec^2 A= 1+cot^2 A=1+(4/3)^2=1+16/9=25/9.
Cosec
A=(25/9)^0.5=+/-5/3.
Cos 16l25
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Sec^2 A= 1+tan^2 A= 1+(3/4)^2=1+9/16=25/16.
Sec A=(25/16)^0.5= +/-5/4.
Cot A=1/tan A=1/(3/4)=4/3.
cosec^2 A= 1+cot^2 A=1+(4/3)^2=1+16/9=25/9.
Cosec
A=(25/9)^0.5=+/-5/3.
Cos 16l25
Mark me brainliest please ❤️
Answered by
2
IN THIS ABOVE QUESTION
TAN A= 3/4
OR, SINA/COSA =3/4
SUPPOSE, SIN A= 3X. AND COS A= 4X.
SECA= 1/ COSA = 1/4X =4X=4.
COSA = 1 / SECA = 1/4.
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