Question

please help me to solve this question​

Answer 1

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Answer 2

Question :

If tanθ+secθ=l,

then prove that secθ =$\dfrac{l{}^{2}+1}{2l}$

Formulas :

• sin²A + cos²A = 1
• sec²A - tan²A = 1
• cosec²A - cot²A = 1

Solution :

we have to prove that secθ =$\dfrac{l{}^{2}+1}{2l}$

Given :

$\tan( \theta) + \sec( \theta) = l$

$\implies \: \tan( \theta) = l - \sec( \theta)$

Now squaring on both sides

$\implies \tan {}^{2} ( \theta) = (l - \sec( \theta) ) {}^{2}$

$\implies \: \tan {}^{2} ( \theta) = l {}^{2} + \sec( \theta) - 2l \sec {}^{2} ( \theta)$

$\implies \: \sec {}^{2} ( \theta) - 1 = l {}^{2} + \sec {}^{2} ( \theta) - 2l \sec( \theta)$

$\implies \: l {}^{2} - 2l \sec( \theta) + 1 = 0$

$\implies \: l {}^{2} + 1 = 2l \sec( \theta)$

$\implies \: \sec( \theta) = \dfrac{l {}^{2} + 1 }{2l}$

$\huge{\bold{ Hence\: Proved}}$

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More Trigonometry Formulas

1. sin2A = 2 sinA cosA
2. cos2A = cos²A - sin²A
3. tan2A = 2 tanA / (1 - tan²A)