Answer
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Hint: Use Pythagorean theorem to find the value of $AB$ and then find the value of $\sin $ and $\cos $. Substitute these values in the given form and use the given conditions to get the exact value.
Complete step by step solution:
Trigonometry ratios are the ratios between edges of the right-angle triangle. There are six trigonometric ratios sin, cos, tan, cosec, sec, cot. Sine function defined as the ratio of perpendicular to the hypotenuse. Cos function is defined as the ratio of base to the hypotenuse. Tan function is defined as the ratio of perpendicular to the base. The reciprocal of these functions defines cosec, sec and cot respectively.
According to the question it is given that in triangle $ABC$,
$BC = 5\;{\text{cm}},\;AC - AB = 1$
To find the value of $AB$ use Pythagorean theorem,
$
A{C^2} = A{B^2} + B{C^2} \\
{\left( {1 + AB} \right)^2} = A{B^2} + {\left( 5 \right)^2} \\
\\
$……..(1)
Now, apply the formula of ${\left( {a + b} \right)^2}$ to find the value of $AB$.
$
\left( {1 + A{B^2} + 2AB} \right) = A{B^2} + 25 \\
\\
$
Cancel out the term $A{B^2}$ from both the sides,
$
2AB = 24 \\
AB = 12{\text{cm}} \\
$
Now, $AC = 1 + AB$ ……(2)
Substitute the value of $AB$ in equation (2),
$
AC = 1 + 12 \\
= 13\;{\text{cm}} \\
$
Here, the value of height $AC = 12\,{\text{cm}}$ and the value of base $BC = 5\;{\text{cm}}$.
Now,
\[
\sin = \dfrac{{{\text{Perpendicular}}}}{{{\text{Hypotenuse}}}} \\
= \dfrac{{12}}{{13}} \\
\]
And,
$
\operatorname{Cos} = \dfrac{{{\text{Base}}}}{{{\text{Hypotenuse}}}} \\
= \dfrac{5}{{13}} \\
$
Thus, substitute the value in the given expression,
$
\dfrac{{1 + \sin C}}{{1 + \cos C}} = \dfrac{{1 + \dfrac{{12}}{{13}}}}{{1 + \dfrac{5}{{13}}}} \\
= \dfrac{{\dfrac{{13 + 12}}{{13}}}}{{\dfrac{{13 + 5}}{{13}}}} \\
= \dfrac{{25}}{{13}} \times \dfrac{{13}}{{18}} \\
= \dfrac{{25}}{{18}} \\
$
Hence, from the above calculation it is concluded that the value of $\dfrac{{1 + \sin C}}{{1 + \cos C}}$ is $\dfrac{{25}}{{18}}$.
Note:Always find the third side by Using Pythagorean theorem in a right-angle triangle and then find the trigonometric ratios. Make sure about the correct formulas of sine and cosine and avoid silly mistakes.
Complete step by step solution:
Trigonometry ratios are the ratios between edges of the right-angle triangle. There are six trigonometric ratios sin, cos, tan, cosec, sec, cot. Sine function defined as the ratio of perpendicular to the hypotenuse. Cos function is defined as the ratio of base to the hypotenuse. Tan function is defined as the ratio of perpendicular to the base. The reciprocal of these functions defines cosec, sec and cot respectively.
According to the question it is given that in triangle $ABC$,
$BC = 5\;{\text{cm}},\;AC - AB = 1$
To find the value of $AB$ use Pythagorean theorem,
$
A{C^2} = A{B^2} + B{C^2} \\
{\left( {1 + AB} \right)^2} = A{B^2} + {\left( 5 \right)^2} \\
\\
$……..(1)
Now, apply the formula of ${\left( {a + b} \right)^2}$ to find the value of $AB$.
$
\left( {1 + A{B^2} + 2AB} \right) = A{B^2} + 25 \\
\\
$
Cancel out the term $A{B^2}$ from both the sides,
$
2AB = 24 \\
AB = 12{\text{cm}} \\
$
Now, $AC = 1 + AB$ ……(2)
Substitute the value of $AB$ in equation (2),
$
AC = 1 + 12 \\
= 13\;{\text{cm}} \\
$
Here, the value of height $AC = 12\,{\text{cm}}$ and the value of base $BC = 5\;{\text{cm}}$.
Now,
\[
\sin = \dfrac{{{\text{Perpendicular}}}}{{{\text{Hypotenuse}}}} \\
= \dfrac{{12}}{{13}} \\
\]
And,
$
\operatorname{Cos} = \dfrac{{{\text{Base}}}}{{{\text{Hypotenuse}}}} \\
= \dfrac{5}{{13}} \\
$
Thus, substitute the value in the given expression,
$
\dfrac{{1 + \sin C}}{{1 + \cos C}} = \dfrac{{1 + \dfrac{{12}}{{13}}}}{{1 + \dfrac{5}{{13}}}} \\
= \dfrac{{\dfrac{{13 + 12}}{{13}}}}{{\dfrac{{13 + 5}}{{13}}}} \\
= \dfrac{{25}}{{13}} \times \dfrac{{13}}{{18}} \\
= \dfrac{{25}}{{18}} \\
$
Hence, from the above calculation it is concluded that the value of $\dfrac{{1 + \sin C}}{{1 + \cos C}}$ is $\dfrac{{25}}{{18}}$.
Note:Always find the third side by Using Pythagorean theorem in a right-angle triangle and then find the trigonometric ratios. Make sure about the correct formulas of sine and cosine and avoid silly mistakes.
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