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Question: In triangle \(ABC\) angle \(B = {90^ \circ }\)and \(BC = 5\;{\text{cm}},\;AC - AB = 1\). Evaluate \(...

In triangle ABCABC angle B=90B = {90^ \circ }and BC=5  cm,  ACAB=1BC = 5\;{\text{cm}},\;AC - AB = 1. Evaluate 1+sinC1+cosC\dfrac{{1 + \sin C}}{{1 + \cos C}}.

Explanation

Solution

Use Pythagorean theorem to find the value of ABAB and then find the value of sin\sin and cos\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 ABCABC,
BC=5  cm,  ACAB=1BC = 5\;{\text{cm}},\;AC - AB = 1
To find the value of ABAB use Pythagorean theorem,
AC2=AB2+BC2 (1+AB)2=AB2+(5)2   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 (a+b)2{\left( {a + b} \right)^2} to find the value of ABAB.
(1+AB2+2AB)=AB2+25   \left( {1 + A{B^2} + 2AB} \right) = A{B^2} + 25 \\\ \\\
Cancel out the term AB2A{B^2} from both the sides,
2AB=24 AB=12cm  2AB = 24 \\\ AB = 12{\text{cm}} \\\
Now, AC=1+ABAC = 1 + AB ……(2)
Substitute the value of ABAB in equation (2),
AC=1+12 =13  cm  AC = 1 + 12 \\\ = 13\;{\text{cm}} \\\
Here, the value of height AC=12cmAC = 12\,{\text{cm}} and the value of base BC=5  cmBC = 5\;{\text{cm}}.
Now,

sin=PerpendicularHypotenuse =1213  \sin = \dfrac{{{\text{Perpendicular}}}}{{{\text{Hypotenuse}}}} \\\ = \dfrac{{12}}{{13}} \\\

And,
Cos=BaseHypotenuse =513  \operatorname{Cos} = \dfrac{{{\text{Base}}}}{{{\text{Hypotenuse}}}} \\\ = \dfrac{5}{{13}} \\\
Thus, substitute the value in the given expression,
1+sinC1+cosC=1+12131+513 =13+121313+513 =2513×1318 =2518  \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 1+sinC1+cosC\dfrac{{1 + \sin C}}{{1 + \cos C}} is 2518\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.