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Basic Trigonometric Identities

Prove that : ...

Question

Prove that : $(sin θ + cos θ) (tan θ + cot θ) = sec θ + cosec θ$

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Solution

Given
$(s i n θ + c o s θ) (t a n θ + c o t θ) = s e c θ + c o s e c θ$

$L . H . S = (s i n θ + c o s θ) (t a n θ + c o t θ)$

$= (s i n θ + c o s θ) (\frac{s i n θ}{c o s θ} + \frac{c o s θ}{s i n θ})$

$= (s i n θ + c o s θ) (\frac{s i n^{2} θ + c o s^{2} θ}{s i n θ . c o s θ})$

$= (s i n θ + c o s θ) \frac{1}{s i n θ . c o s θ}$

$= \frac{s i n θ + c o s θ}{s i n θ . c o s θ}$

$= \frac{s i n θ}{s i n θ c o s θ} + \frac{c o s θ}{s i n θ c o s θ}$

$= \frac{1}{c o s θ} + \frac{1}{s i n θ}$

$= s e c θ + c o s e c θ$

$= R . H . S$

$∴ (s i n θ + c o s θ) (t a n θ + c o t θ) = s e c θ + c o s e c θ$

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Similar questions

(a) Prove that $\frac{c o s A}{1 + s i n A} + t a n A = s e c A$

(b) Prove the identity $(s i n θ + c o s θ) (t a n θ + c o t θ) = s e c θ + c o s e c θ$ [6 MARKS]

Prove the following trigonometric identities.
(i) $\frac{1 + cos θ + sin θ}{1 + cos θ - sin θ} = \frac{1 + sin θ}{cos θ}$

(ii) $\frac{sin θ - cos θ + 1}{sin θ + cos θ - 1} = \frac{1}{sec θ - tan θ}$

(iii) $\frac{cos θ - sin θ + 1}{cos θ + sin θ - 1} = cosec θ + cot θ$

(iv) $(sin θ + cos θ) (tan θ + cot θ) = sec θ + cosec θ$

(cosec θ - sin θ) (sec θ - cos θ) (tan θ + cot θ) = . . . . . . .

Q. Solve:

(sin θ + cos θ) (tan θ + cot θ) = csc θ + cot θ

Q. Evaluate

(c o s e c θ - sin θ) (sec θ - cos θ) (tan θ + cot θ)

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Prove that : (sinθ+cosθ)(tanθ+cotθ)=secθ+cosecθ

Prove that : $(sin θ + cos θ) (tan θ + cot θ) = sec θ + cosec θ$