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Trigonometric Ratios

Question 7 ii...

Question

Question 7 (ii)
If $c o t θ$ = $\frac{7}{8}$ , evaluate:
(ii) $c o t^{2} θ$

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Solution

(ii) $c o t^{2} θ = {(\frac{7}{8})}^{2} = \frac{49}{64}$

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\frac{c o t θ - c o t 2 θ}{c o t θ}

c o t^{2} θ + c o t (θ - α) c o t (θ - β)

c o t 2 θ = \frac{1}{2} (c o t α + c o t β)

c o t θ = \frac{7}{8}

\frac{1 + s i n θ}{c o s θ}

Q. Prove that :

\frac{t a n θ - c o t θ}{s i n θ c o s θ} = s e c^{2} θ - c o s e c^{2} θ = t a n^{2} θ - c o t^{2} θ

t a n θ + c o t θ = 2

, find the value of

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Trigonometric Ratios

Standard X Mathematics

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