Question

a square sec square theta minus b square tan square theta = c square then prove that sin theta equals to plus minus under root c square minus a square upon c square minus b square

Answer 1

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Good luck

Answer 2

Answer:

$sin\theta =±\left(\sqrt{ \frac{c^{2}-a^{2}}{a^{2}-b^{2}}}\right)$

Step-by-step explanation:

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

$\implies a^{2}sec^{2}\theta-b^{2}(sec^{2}\theta-1)=c^{2}$

/*By Trigonometric identity:

tan²A=sec²A-1 */

$\implies a^{2}sec^{2}\theta-b^{2}-sec^{2}\theta+b^{2}=c^{2}$

$\implies (a^{2}-b^{2})sec^{2}\theta=c^{2}-b^{2}$

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

$\implies cos^{2}\theta = \frac{a^{2}-b^{2}}{c^{2}-b^{2}}$

$\implies 1- sin^{2}\theta = \frac{a^{2}-b^{2}}{c^{2}-b^{2}}$

$\implies sin^{2}\theta= 1-\frac{a^{2}-b^{2}}{c^{2}-b^{2}}$

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

Therefore,

$sin\theta =±\left(\sqrt{ \frac{c^{2}-a^{2}}{a^{2}-b^{2}}}\right)$

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