Question

If sinQ=12/13,find the value of (sin^2Q-cos^2Q/sin^2Q+cos^2Q)×1/tan^2Q

Answer 1

answer...............

Answer 2

Answer:

$\text{The value is }\frac{2975}{24336}$

Step-by-step explanation:

Given $sinQ=\frac{12}{13}$

we have to find the value of $\frac{(sin^2Q-cos^2Q)}{(sin^2Q+cos^2Q)}\times\frac{1}{tan^2Q}$

$sinQ=\frac{12}{13}$

$sin^2Q+cos^2Q=1$

⇒ $(\frac{12}{13})^2+cos^2Q=1$

⇒ $cos^2Q=1-\frac{144}{169}=\frac{25}{169}$

Now, $\frac{(sin^2Q-cos^2Q)}{(sin^2Q+cos^2Q)}\times\frac{1}{tan^2Q}$

$=\frac{((\frac{12}{13})^2-(\frac{5}{13})^2)}{1}\times\frac{cos^2Q}{sin^2Q}\\\\=(\frac{144}{169}-\frac{25}{169})\times\frac{(\frac{5}{13})^2}{(\frac{12}{13})^2}\\\\=\frac{119}{169}\times\frac{25}{144}\\\\=\frac{2975}{24336}$