Question

prove that 10 square theta minus sin square theta is equal to 10 square theta into sin square theta

Answer 1

here is ur answer.....

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Answer 2

Answer:

$tan^{2}\theta-sin^{2}\theta =sin^{2}\theta tan^{2}\theta$

Step-by-step explanation:

$LHS = tan^{2}\theta-sin^{2}\theta \\=\frac{sin^{2}\theta}{cos^{2}\theta}-sin^{2}\theta}\\=sin^{2}\theta\left(\frac{1}{cos^{2}\theta}-1}\right)\\=sin^{2}\theta\left(\frac{1-cos^{2}\theta}{cos^{2}\theta}\right)\\=sin^{2}\theta \left(\frac{sin^{2}\theta}{cos^{2}\theta}\right)\\=sin^{2}\theta tan^{2}\theta\\=RHS$

Therefore,

$tan^{2}\theta-sin^{2}\theta =sin^{2}\theta tan^{2}\theta$

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