Prove the ideintity- -1-tan-1-tan-1-sin-cos-4 MARKS-

Formula: 2 Marks Proof: 2 Marks LHS=(sec θ+tan θ)+1/se θ+1 tan θ=(sec θ+tan θ)+(sec^2 θ tan^2 θ)/(sec θ+1 tan θ) =(sec θ+tan θ)+(sec θ tan θ)(sec θ+tan θ)/se ...

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Standard X

Mathematics

Trigonometric Identities

Prove the ide...

Question

Prove the ideintity: $\frac{s e c θ + 1 + t a n θ}{s e c θ + 1 - t a n θ} = \frac{1 + s i n θ}{c o s θ}$ [4 MARKS]

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Solution

Formula: 2 Marks
Proof: 2 Marks

LHS $= \frac{(s e c θ + t a n θ) + 1}{s e θ + 1 - t a n θ} = \frac{(s e c θ + t a n θ) + (s e c^{2} θ - t a n^{2} θ)}{(s e c θ + 1 - t a n θ)} = \frac{(s e c θ + t a n θ) + (s e c θ - t a n θ) (s e c θ + t a n θ)}{s e c θ + 1 - t a n θ} [∵ s c^{2} θ - t a n^{2} θ = 1 = (s e c θ - t a n θ) (s e c θ + t a n θ)] = \frac{(s e c θ + t a n θ) (1 + s e c θ - t a n θ)}{(s e c θ + 1 - t a n θ)} = s e c θ + t a n θ = \frac{1}{c o s θ} + \frac{s i n θ}{c o s θ} = \frac{1 + s i n θ}{c o s θ} = R H S$

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Similar questions

Q. Prove that

\frac{s e c θ + t a n θ - 1}{t a n θ - s e c θ + 1} = \frac{c o s θ}{(1 - s i n θ)}

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Q. Question 12
Prove that

\frac{1 + s e c θ - t a n θ}{1 + s e c θ + t a n θ} = \frac{1 - s i n θ}{c o s θ}

Prove the following

1. $(1 - {sin}^{2} A) s e c^{2} A = 1$

2. ${sec}^{4} θ - {sec}^{2} θ = {tan}^{4} θ + {tan}^{2}$

3. $(sec θ - tan θ)^{2} = \frac{1 - sin θ}{1 + sin θ}$

4. $\frac{tan θ + sec θ - 1}{tan θ - sec θ +} = \frac{1 + sin θ}{cos θ}$

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Prove the ideintity: sec θ+1+tan θsec θ+1−tan θ=1+sin θcos θ [4 MARKS]

Prove the ideintity: $\frac{s e c θ + 1 + t a n θ}{s e c θ + 1 - t a n θ} = \frac{1 + s i n θ}{c o s θ}$ [4 MARKS]