Question

A+B=135°, then (1+cotA)(1+cotB)​

Answer 1

A+B=135°

cot(A+B)=cot(135°)

cotAcotB-1=-cotA-cotB

1+cotA+cotB+cotAcotB=2

1(1+cotA)+cotB(1+cotA)=2

(1+cotA)(1+cotB)=2

Answer 2

Answer:

0

Step-by-step explanation:

Given :

A + B = 135° ,.

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To Find :

The value of : (1+cot A)(1+cot B)

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Solution:

If A + B = 135°

⇒ A = 135° - B ---> (i)

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(1 + cot A)(1 + cot B)

⇒ (1 + cot (135 - B))(1 + cot B)

∴ $tan(X-Y) = \frac{tan X - tan Y}{1+tanXtanY$

⇒ $\frac{1}{tan(X-Y)} = cot(X-Y) = \frac{1+tanXtanY}{tanX - tanY}$}

Applying the formula, X = 135°,Y=B

We get,

⇒ $(1+\frac{1+tan135tanB}{-1-tan135tanB}) (1 + cotB)$     ∴ Tan 135° = -1

⇒  $(1+\frac{1+tan135tanB}{-(1+tan135tanB)}) (1 + cotB)$

⇒ $(1+(-1))(1+cotB)$ = $(0)(1+cotB) = 0$

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Hope it Helps!