Question

If "a sin square theta + b cos square theta" equal to c.Then what is the value of tan square theta????

Answer 1

$a \ { \sin}^{2} \alpha + b { \cos}^{2} \alpha = c$
${ \tan}^{2} \alpha = \frac{(c - b { \ \cos }^{2} \alpha )b }{(c - {a \sin}^{2} \alpha ) a}$
${ \tan }^{2} \alpha = \frac{cb - {b}^{2} { \cos}^{2} \alpha }{ca - {a}^{2} { \sin}^{2} \alpha }$

Answer 2

Answer:

the value of  $\tan^{2} \theta$ is $\frac{c \sec^{2} \theta-b}{a}$

Step-by-step explanation:

If $a \sin^{2} \theta +b \cos^{2} \theta = c$

we need to find the value of $\tan^{2} \theta$

As,  $a \sin^{2} \theta +b \cos^{2} \theta = c$

divide both the side by  $a \cos^{2} \theta$

$\frac{a \sin^{2} \theta}{a \cos^{2} \theta} +\frac{b \cos^{2} \theta}{{a \cos^{2} \theta}} = \frac{c}{{a \cos^{2} \theta}}$

$\tan^{2} \theta+\frac{b}{a} =\frac{c}{a}\sec^{2} \theta$

subtract both the sides by $\frac{b}{a}$

$\tan^{2} \theta =\frac{c}{a}\sec^{2} \theta - \frac{b}{a}$

$\tan^{2} \theta =\frac{c\sec^{2} \theta-b}{a}$

Hence, the value of  $\tan^{2} \theta$ is $\frac{c \sec^{2} \theta-b}{a}$