1-tan -cot - sin -cos - - sec -cosec2 -cosec -sec2 -

LHS = (1+tan θ+cot θ) (sin θ cos θ) =(1+ sin θ/cos θ + cos θ/sin θ ) (sin θ cos θ) = ( 1 + sin^2 θ + cos^2 θ/sin θ cos θ)(sin θ cos θ) = ( 1 + 1/sin θ ...

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Byju's Answer

Standard X

Mathematics

Trigonometric Identities

1+tan θ+cot θ...

Question

$(1 + t a n θ + c o t θ) (s i n θ - c o s θ) = (\frac{s e c θ}{c o s e c^{2} θ} - \frac{c o s e c θ}{s e c^{2} θ})$

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Solution

$L H S = (1 + t a n θ + c o t θ) (s i n θ - c o s θ) = (1 + \frac{s i n θ}{c o s θ} + \frac{c o s θ}{s i n θ}) (s i n θ - c o s θ) = (1 + \frac{s i n^{2} θ + c o s^{2} θ}{s i n θ c o s θ}) (s i n θ - c o s θ) = (1 + \frac{1}{s i n θ c o s θ}) (s i n θ - c o s θ) = \frac{(1 + s i n θ c o s θ) (s i n θ - c o s θ)}{s i n θ c o s θ} R H S = (\frac{s e c θ}{c o s e c^{2} θ} - \frac{c o s e c θ}{s e c^{2} θ}) = \frac{s i n^{2} θ}{c o s θ} - \frac{c o s^{2} θ}{s i n θ} = \frac{s i n^{3} θ - c o s^{3} θ}{s i n θ c o s θ} = \frac{(s i n θ - c o s θ) (s i n^{2} θ + c o s^{2} θ + s i n θ c o s θ)}{s i n θ c o s θ} [∵ a^{3} - b^{3} = (a - b) (a^{2} + b + a b)] = \frac{(1 + s i n θ c o s θ) (s i n θ - c o s θ)}{s i n θ c o s θ} L H S = R H S$

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(1+tanθ+cotθ)(sinθ−cosθ)=(secθcosec2θ−cosecθsec2θ)

$(1 + t a n θ + c o t θ) (s i n θ - c o s θ) = (\frac{s e c θ}{c o s e c^{2} θ} - \frac{c o s e c θ}{s e c^{2} θ})$