Write all the trigonometric ratios of $\angle A$ in terms of $\sec A$

Hint:We have to find how $\sec A$ is related to the other trigonometric functions.Using trigonometric identities and rearranging the terms, we will get the trigonometric ratios of $\cos A,\sin A,\tan A,\cot A$ and $\cos ecA$ in terms of $\sec A$.

Complete step-by-step answer:
Now we check how all trigonometric ratios are related to $\sec A$
(i) $\cos A$
As we know that $\cos A$ and $\sec A$ are related to each other by the equation, which means that they are reciprocal to each other, secant is the reciprocal of the cosecant.
$\cos A$$ = \dfrac{1}{{\sec A}}$

(ii) $\tan A$
As we know that $\sec A$ an $\tan A$ are related by the equation,
$1 + {\tan ^2}A = {\sec ^2}A$
Rearranging the terms, we get,
${\tan ^2}A = 1 - {\sec ^2}A$
Taking the square root on both side of the equation,
$\tan A = \pm \sqrt {{{\sec }^2}A - 1} $
Here, we know that $A$ is acute that means it is less than $90$ degrees and $\tan A$ is positive when $A$ is acute.
Therefore,
$\tan A = \sqrt {{{\sec }^2}A - 1} $

(iii) $\cot A$
As we know that tangent and cotangent are reciprocal to each other.
So, we can write $\cot A$ as,
$\cot A = \dfrac{1}{{\tan A}}$
Substitute the value of $\tan A = \sqrt {{{\sec }^2}A - 1} $ in the above equation then we get,
$\cot A = \dfrac{1}{{\sqrt {{{\sec }^2}A - 1} }}$

(iv) $\cos ecA$
As we know that cosecant and cotangent are related by the relation,
$1 + {\cot ^2}A = \cos e{c^2}A$
Rearranging the terms so we get,
$\cos e{c^2}A = 1 + {\cot ^2}A$
Substitute the value of $\cot A$ ,
$\cos e{c^2}A = 1 + {\left( {\dfrac{1}{{\sqrt {{{\sec }^2}A - 1} }}} \right)^2}$
Cancelling the square root ,
$\cos e{c^2}A = 1 + \left( {\dfrac{1}{{{{\sec }^2}A - 1}}} \right)$
Cross multiplying the above expression to get,
$\cos e{c^2}A = \left( {\dfrac{{1 \times \left( {{{\sec }^2}A - 1} \right) + 1}}{{{{\sec }^2}A - 1}}} \right)$
Expanding the bracket,
$
  \cos e{c^2}A = \left( {\dfrac{{{{\sec }^2}A - 1 + 1}}{{{{\sec }^2}A - 1}}} \right) \\
   = \dfrac{{{{\sec }^2}A}}{{{{\sec }^2}A - 1}} \\
 $
Then taking the square roots on both sides,
$\cos ecA = \pm \sqrt {\dfrac{{{{\sec }^2}A}}{{{{\sec }^2}A - 1}}} $
Taking square root on numerator to get,
$\cos ecA = \pm \dfrac{{\sec A}}{{\sqrt {{{\sec }^2}A - 1} }}$
Here, we know that $A$ is acute that means it is less than $90$ degrees and is positive when $A$ is acute.
Therefore,
$\cos ecA = \dfrac{{\sec A}}{{\sqrt {{{\sec }^2}A - 1} }}$

(v) $\sin A$
As we know that sine and cosecant are reciprocal to each other, hence we can write
$\sin A = \dfrac{1}{{\cos ecA}}$
Substitute the value of $\cos ecA = \dfrac{{\sec A}}{{\sqrt {{{\sec }^2}A - 1} }}$ in the above equation then we get,
$\sin A = \dfrac{1}{{\left( {\dfrac{{\sec A}}{{\sqrt {{{\sec }^2}A - 1} }}} \right)}}$
Taking the reciprocal, we get,
$\sin A = \dfrac{{\sqrt {{{\sec }^2}A - 1} }}{{\sec A}}$

Hence, we got all the trigonometric ratios of $\angle A$ in terms of $\sec A$

Note:In order to find the trigonometric ratios of $\angle A$ in terms of $\sec A$ ,we need to find how each trigonometric function related to each other, hence we will get their relation with each other in terms of any trigonometric function.Students should remember trigonometric identities i.e $1 + {\tan ^2}A = {\sec ^2}A$, $1 + {\cot ^2}A = \cos e{c^2}A$ and reciprocal identities $\sin A = \dfrac{1}{{\cos ecA}}$, $\cos A$$ = \dfrac{1}{{\sec A}}$, $\cot A = \dfrac{1}{{\tan A}}$ for solving these types of problems.