Prove the identity-i A -tan B - B -tan A - A tan Bii 1-tan 2-1- 2-1-tan-1-2 -4 MARKS-

Each Part: 2 Marks (i) LHS=cot A+tan B/cot B+tan A= ( 1/tan A )+tan B/ ( 1/tan B )+tan A =(1+tan A tan B)/tan A×tan B/(1+tan A tan B) =cot A tan B=RHS (ii)L ...

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Standard X

Mathematics

Trigonometric Ratios

Prove the ide...

Question

Prove the identity:
(i) $\frac{c o t A + t a n B}{c o t B + t a n A} = c o t A t a n B$
(ii) $\frac{1 + t a n^{2} θ}{1 + c o t^{2} θ} = {(\frac{1 - t a n θ}{1 - c o t θ})}^{2}$ [4 MARKS]

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Solution

Each Part: 2 Marks

(i) $L H S = \frac{c o t A + t a n B}{c o t B + t a n A} = \frac{(\frac{1}{t a n A}) + t a n B}{(\frac{1}{t a n B}) + t a n A} = \frac{(1 + t a n A t a n B)}{t a n A} \times \frac{t a n B}{(1 + t a n A t a n B)} = c o t A t a n B = R H S$

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

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[3 MARKS]

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Prove the identity: (i) cot A+tan Bcot B+tan A=cot A tan B (ii) 1+tan2 θ1+cot2 θ=(1−tan θ1−cot θ)2 [4 MARKS]

Prove the identity:
(i) $\frac{c o t A + t a n B}{c o t B + t a n A} = c o t A t a n B$
(ii) $\frac{1 + t a n^{2} θ}{1 + c o t^{2} θ} = {(\frac{1 - t a n θ}{1 - c o t θ})}^{2}$ [4 MARKS]