Question

Solve the given expression as,
$\cot A+\csc A=5$ . Find the value of $\cos A$ .

Answer 1

Hint: Using bas identities convert the given expression in terms of $\cos A$ and $\sin A$. Equate the expression to k and find the expression for $\cos A$ and $\sin A$ . Equate them in ${{\cos }^{2}}A+{{\sin }^{2}}A=1$ and solve for k. Thus, substitute the value of k back and get the value of $\cos A$ .

Complete step-by-step answer:
We have been given an expression from which we have to find the value of $\cos A$ .
We have been given, $\cot A+\csc A=5$ ………………………………….(1)
We know the basic identity that $\cot A=\dfrac{\cos A}{\sin A}$ and $\csc A=\dfrac{1}{\sin A}$
Thus, we can rewrite (1) as,
$\cot A+\csc A=5$
$\Rightarrow \dfrac{\cos A}{\sin A}+\dfrac{1}{\sin A}=5$
$\therefore \dfrac{1+\cos A}{\sin A}=5$ , now let us apply cross multiplication property
$\dfrac{1+\cos A}{5}=\sin A=k$ , let us say that its equal to k
$\therefore \dfrac{1+\cos A}{5}=k$ and $\sin A=k$
$\therefore \cos A=5k-1$ and $\sin A=k$
We know that ${{\cos }^{2}}\theta +{{\sin }^{2}}\theta =1$
Similarly, ${{\cos }^{2}}A+{{\sin }^{2}}A=1$
Apply values of $\cos A$ and $\sin A$ in the above expression.
${{\cos }^{2}}A+{{\sin }^{2}}A=1$
${{\left( 5k-1 \right)}^{2}}+{{k}^{2}}=1$
We know the basic identity, ${{\left( a-b \right)}^{2}}={{a}^{2}}-2ab+{{b}^{2}}$ . We can expand the above as,
${{\left( 5k \right)}^{2}}-2\times 5k\times 1+{{1}^{2}}+{{k}^{2}}=1$
$25{{k}^{2}}-10k+1+{{k}^{2}}=1$
i.e. $26{{k}^{2}}-10k=0$
$2k\left( 13k-5 \right)=0$
Thus, we get that $2k=0$ and $13k-5=0$ $\Rightarrow k=\dfrac{5}{13}$
Hence, we get the value of $k=0$ and $k=\dfrac{5}{13}$
Now, $\cos A=5k-1$
Let us put the value of k as 0 and $\dfrac{5}{13}$ in the above expression when $k=0$ , $\cos A=5\times 0-1$
$\therefore \cos A=-1$
when $k=\dfrac{5}{13},\text{ }\cos A=\dfrac{5\times 5}{13}-1$
$=\dfrac{25-13}{13}=\dfrac{12}{13}$
$\therefore \cos A=\dfrac{12}{13}$
Hence, we got the value of $\cos A$ as $-1$ and $\dfrac{12}{13}$ .

Note: We have given that the expression is equal to a numeral 5. Thus, we need a numeral or fraction as the value of $\cos A$ . Don’t stop it as $\cos A=5\sin A-1$ , as the value of $\cos A$ . Equate the expression to k and get the value of k.