Question

Prove that 1 sin theta divided by 1- sin theta minus 1-sin theta divided by 1 sin theta =4 tan theta × sec theta

Answer 1

please go through the question thoroughly

Step-by-step explanation:

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Answer 2

$LHS = \frac{1+sin\theta}{1-sin\theta} - \frac{1-sin\theta}{1+sin\theta}\\= \frac{ (1+sin\theta)^{2} - (1-sin\theta)^{2}}{(1-sin\theta)(1+sin\theta)}\\= \frac{4 \times 1 \times sin\theta}{ 1^{2} - sin^{2} \theta}$

$\underline { \pink { By \: following \: identities : }}$

• $\blue { (a+b)^{2} - (a-b)^{2} = 4ab}$
• $\blue { (a+b)(a-b) = a^{2} - b^{2} }$

$= \frac{4 sin \theta}{ cos^{2}\theta}$

$\boxed { \orange { Since, 1 - sin^{2} \theta = cos^{2}\theta }}$

$= 4 \times \frac{sin \theta}{cos\theta} \times \frac{1}{cos\theta} \\= 4 tan \theta sec \theta \\= RHS$

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