Question

tan2A = cot(A-18),then sec( A+9) =​

Answer 1

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Answer 2

Answer:

The value of $sec(A+9)$ is $\sqrt{2}$.

Step-by-step explanation:

Here we have given that, it this equation we will find the value of A by equation the equation on the left-hand side and right-hand side. Then we will substitute the value of A in $sec(A+9)$, and find the value of  $sec(A+9)$.

$tan2A=cot(A-18)$

We will use the trigonometry equation which is,

$tan\theta=cot(90^ ^{\circ} -\theta)$

$cot(90-2A)=cot(A-18)$

$90-2A=A-18$

$3A=108$$A=\frac{108}{3}$

$A=36$

Now, the finding the value of $sec(A+9)$. Therefore, substituting the value of A in $sec(A+9)$.

$sec(A+9)=sec(36+9)\\sec(A+9)=sec45^ ^{\circ}$

The value of $sec45$ is $\sqrt{2}$.