If tan A-tan B-a and A- B-b- prove that- A-B-1-a-1-b-

We have, tanA+tanB=a and cotA+cotB=b Now, cotA+cotB=b ⇒1/tanA+1/tanB=b ⇒tanB+tanA/tanAtanB =b ⇒a/b =tanAtanB ∵ cot(A+B)=1/tan(A+B) = 1/tanA+tanB/1 tanAtanB = 1 ...

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Byju's Answer

Standard XII

Mathematics

Trigonometric Ratios for Sum of Two Angles

If tan A+tan ...

Question

If tanA+tanB=a and cotA+ cotB=b, prove that: cot(A+B)= $\frac{1}{a} - \frac{1}{b}$ .

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Solution

We have,
tanA+tanB=a and cotA+cotB=b
Now, cotA+cotB=b
$\Rightarrow \frac{1}{t a n A} + \frac{1}{t a n B} = b$
$\Rightarrow \frac{t a n B + t a n A}{t a n A t a n B} = b$
$\Rightarrow \frac{a}{b} = t a n A t a n B$
$∵ c o t (A + B) = \frac{1}{t a n (A + B)}$
$= \frac{\frac{1}{t a n A + t a n B}}{1 - t a n A t a n B}$
$= \frac{1 - \frac{a}{b}}{a}$
$= \frac{b - a}{a b}$
$= \frac{b}{a b} - \frac{a}{a b}$
$= \frac{1}{a} - \frac{1}{b}$
$∴ c o t (A + B) = \frac{1}{a} - \frac{1}{b}$
Hence proved.

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If tanA+tanB=a and cotA+ cotB=b, prove that: cot(A+B)= 1a−1b.

We have, tanA+tanB=a and cotA+cotB=b Now, cotA+cotB=b ⇒1tanA+1tanB=b ⇒tanB+tanAtanAtanB=b ⇒ab=tanAtanB ∵cot(A+B)=1tan(A+B) =1tanA+tanB1−tanAtanB =1−aba =b−aab =bab−aab =1a−1b ∴cot(A+B)=1a−1b Hence proved.

If tanA+tanB=a and cotA+ cotB=b, prove that: cot(A+B)= $\frac{1}{a} - \frac{1}{b}$ .