Question

19. In a right ABC, angle B=90° AC-BC=2. AB= 4√2. Then sec A+cotc?
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Answer 1

the question,

We have a right angled triangle ABC at B.

AC = 2√5

AB - BC = 2

We need to find the value of,

cos²A - cos²C

So,

We can say,

AB = x + 2

BC = x

In the triangle using Pythagoras Theorem,

AC² = AB² + BC²

So,

\begin{gathered}(2\sqrt{5})^{2}=(x+2)^{2}+(x)^{2}\\20=2x^{2}+4+4x\\2x^{2}+4x=16\\x^{2}+2x-8=0\\(x+4)(x-2)=0\\x=-4,2\end{gathered}

(2

5

)

2

=(x+2)

2

+(x)

2

20=2x

2

+4+4x

2x

2

+4x=16

x

2

+2x−8=0

(x+4)(x−2)=0

x=−4,2

So,

x = 2

Now,

AB = 4

and,

BC = 2

So,

\begin{gathered}cos^{2}A=(\frac{4}{2\sqrt{5}})^{2}\\cos^{2}A=\frac{4}{5}\end{gathered}

cos

2

A=(

2

5

4

)

2

cos

2

A=

5

4

also.\begin{gathered}cos^{2}C=(\frac{2}{2\sqrt{5}})^{2}\\cos^{2}C=\frac{1}{5}\end{gathered}

cos

2

C=(

2

5

2

)

2

cos

2

C=

5

1

So,\begin{gathered}cos^{2}A-cos^{2}C=\frac{4}{5}-\frac{1}{5}\\cos^{2}A-cos^{2}C=\frac{3}{5}\end{gathered}

cos

2

A−cos

2

C=

5

4

−

5

1

cos

2

A−cos

2

C=

5

3

Therefore, the required value is,

\frac{3}{5}

5

3