I If A - B - 45o- prove that cot A-1cotB-1-2 and hence deduce that cot 22 1-2o - -2 - 1-ii If tanA-B - 7-24 and tan A-4-3 where A and B are acute- show that A - B - -2

(i) cot(A+B)=cotA.cotB 1cotA+cotB and A+B=45 = gt; cotA.cotB 1=cotA+cotB = gt; cotA.cotB cotA cotB 1+2=2 = gt; (cotA 1)(cotB 1)=2 ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XII

Mathematics

Principal Solution of Trigonometric Equation

i If A + B = ...

Question

(i) If A + B = $45^{o}$ , prove that $(c o t A - 1) (c o t B - 1) = 2$ and hence deduce that $c o t 22 {\frac{1}{2}}^{o} = \sqrt{2} + 1$ .
(ii) If $t a n (A - B) = \frac{7}{24}$ and $t a n A = \frac{4}{3}$ where A and B are acute, show that A + B = $\frac{π}{2}$

Open in App

Solution

(i) $c o t (A + B) = \frac{c o t A . c o t B - 1}{c o t A + c o t B}$
and $A + B = 45$
=> $c o t A . c o t B - 1 = c o t A + c o t B$
=> $c o t A . c o t B - c o t A - c o t B - 1 + 2 = 2$
=> $(c o t A - 1) (c o t B - 1) = 2$

If $A = B = \frac{π}{8}$ , and $c o t \frac{π}{8} = x$

=> $(x - 1) (x - 1) = 2$
=> $x^{2} - 2 x - 1 = 0$
=> $x = \sqrt{2} + 1$

(ii) $t a n (A - B) = \frac{t a n A - t a n B}{1 + t a n A . t a n B}$
=> $\frac{7}{24} = \frac{\frac{4}{3} - t a n B}{1 + \frac{4}{3} t a n B}$
=> simplifying, we get $t a n B = \frac{3}{4}$

Now, $t a n (A + B) = \frac{t a n A + t a n B}{1 - t a n A . t a n B}$

$1 - t a n A . t a n B = 1 - \frac{4}{3} . \frac{3}{4} = 0$
=> $A + B = \frac{π}{2}$

Suggest Corrections

Similar questions

If A and B are acute angles such that tan $A = \frac{1}{2}, t a n B = \frac{1}{3}$ and tan $(A + B) = \frac{t a n A + t a n B}{1 - t a n A t a n B}$ ,find A + B

(i) If $t a n A = \frac{5}{6}$ and $t a n B = \frac{1}{11},$ prove that A+B= $\frac{π}{4}$
(ii) If $t a n A = \frac{m}{m - 1}$ and $t a n B = \frac{m}{2 m - 1}$
then prove that $A - B = \frac{π}{4}$

Q. If

tan A = cot B

, prove that

A + B + 90

, where A and B are measures of acute angles.

Q. If tan(A + B) =

\sqrt 3

and tan(A - B) =

\frac{1}{\sqrt{3}} .

Find A and B where A & B are acute angles.

If tan(A+B) = $\sqrt{3}$ and sin(A-B) = $\frac{1}{2}$ Where A and B are acute angles.

Then, Cos2A + Sin2B =__

Join BYJU'S Learning Program

(i) If A + B = 45o, prove that (cotA−1)(cotB−1)=2 and hence deduce that cot2212o=√2+1.(ii) If tan(A−B)=724 and tanA=43 where A and B are acute, show that A + B = π2

(i) If A + B = $45^{o}$ , prove that $(c o t A - 1) (c o t B - 1) = 2$ and hence deduce that $c o t 22 {\frac{1}{2}}^{o} = \sqrt{2} + 1$ .
(ii) If $t a n (A - B) = \frac{7}{24}$ and $t a n A = \frac{4}{3}$ where A and B are acute, show that A + B = $\frac{π}{2}$