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Question 4sin...

Question

Question 4
$(s i n α + c o s α) (t a n α + c o t α) = s e c α + c o s e c α$

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Solution

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

$= (s i n α + c o s α) (\frac{s i n α}{c o s α} + \frac{c o s α}{s i n α})$ $[∵ t a n θ = \frac{s i n θ}{c o s θ} a n d c o t θ = \frac{c o s θ}{s i n θ}]$

$= (s i n α + c o s α) (\frac{s i n^{2} α + c o s^{2} α}{s i n α . c o s α})$ $[∵ s i n^{2} θ + c o s^{2} θ = 1]$

$= \frac{1}{c o s α} + \frac{1}{s i n α}$ $[∵ s e c θ = \frac{1}{c o s θ} a n d c o s e c θ = \frac{1}{s i n θ}]$

$= s e c α + c o s e c α$

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Similar questions

Q. Question 6

1 + \frac{c o t^{2} α}{1 + c o s e c α} = c o s e c α

cot α + tan α = m

\frac{1}{cos α} - cos α = n

α

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cos α = cot α

Q. The values of

c o s α, t a n α, c o t α,

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s i n α = \frac{5}{13}

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