Question

In a ΔABC right triangle at B. If $SinC=\frac{60}{61}$ and $CosA=\frac{60}{61}$, then find the value of the expression $\frac{tanA+cot(A+C)}{CotC+SecASinC}$

• A. 11/61
• B. 11/60
• C. 11/71
• D. 11/21

Answer 1

Answer:

C. 11/71

Solution:

Given:

In a ΔABC right triangle at B.

$SinC=\frac{60}{61}$

$CosA=\frac{60}{61}$

Concept used:

$Sin\theta=\frac{p}{h}=\frac{60}{61},Cos\theta=\frac{b}{h}$

If x + y = 90^∘, then sinx = siny

Calculation:

$SinC=\frac{p}{h}=\frac{60}{61}$

The value of base 'b' for the triangle ΔABC right-angle at B can be calculated by using the Pythagoras theorem.

⇒ b² = h² - p²

⇒ b² = 61² - 60²

⇒ b² = (61 + 60)(61 - 60)

⇒ b² = 121

⇒ b = 11 units

According to the question, the required image is:

Now, 

$tanA=\frac{11}{60}$

$CotC=\frac{11}{60}$

$SecA=\frac{61}{60}$

Since $sinC=cosA=\frac{60}{61}$

A + C = 90^∘

cot(A + C) = cot(90^∘) = 0

Now, substitute the value of the respective variable in the required expression,

$=\frac{\frac{11}{60}+0}{\frac{11}{60}+\frac{61}{60}\times{\frac{60}{61}}}$

$=\frac{\frac{11}{60}}{\frac{11}{60}+1}$

$=\frac{\frac{11}{60}}{\frac{71}{60}}$

$=\frac{11}{71}$

The value of the required expression is$=\frac{11}{71}$