Question

if (sec A-2)(tan3A-1) = 0 find a​

Answer 1

Answer:

$A = 15° , 60° , 75°  are \: the \: solution$

Step-by-step explanation:

Solve the equation (sec A-2)(tan 3A-1)=0

(sec A-2) = 0

=> secA - 2 = 0

=> secA = 2

=> 1/CosA = 2

=> 2CosA = 1

=> CosA = 1/2

=> A = 60°

tan 3A-1 = 0

=> tan3A = 1

=> 3A = 45° or 225°

=> A = 45° /3 or 225°/3

=> A = 15° or 75°

A = 15° , 60° , 75° are the solution

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Answer 2

According to the question .

Step by step explaination

Sec A - 2 = 0

(Sec A = 1/Cos A)

1/CosA = 2

2 CosA =1

Cos A =1/2

A = 60°

Tan 3 A=1.

3A=1/Tan

A = 45/3

A= 15°

A CAN BE 60° , 15° , 75°.

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