Question

If cosA= 3/5 , find 9 cot^2 A-1.

Answer 1

cosA = 3/5
sinA = 4/5
cotA = 3/4
9 cot²A – 1 = 9(3/4)² – 1 = 9×9/16 – 1 = 65/16

Answer 2

Answer:

$9cot^{2}A-1=\frac{65}{16}$

Step-by-step explanation:

$We \:have ,\: \\cosA=\frac{3}{5}--(1)$

\* We know the Trigonometric identity:

sin²A = 1-cos²A *\

$\implies sin^{2}A\\=1-(\frac{3}{5})^{2}\\=1-\frac{9}{25}\\=\frac{25-9}{25}\\=\frac{16}{25}--(2)$

Now,

$9cot^{2}A-1\\=9\big(\frac{cos^{2}A}{sin^{2}A}\big)-1\\=9\big(\frac{\frac{9}{25}}{\frac{16}{25}}\big)-1$

/* from (1)&(2)*\

$=9\times \frac{9}{16}-1$

$=\frac{81}{16}-1\\=\frac{81-16}{16}\\=\frac{65}{16}$

Therefore,

$9cot^{2}A-1=\frac{65}{16}$

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