I If tan A -5-6 and tan B -1-11- prove that A - B -4ii If tan A - m - m -1 and tan B - m -2 m -1then prove that A - B -4

We have tanA= 5/6 and tanB= 1/11 Now, tan(A+B)= tanA+tanB/1 tanA.tanB =5/6+1/11/1 5/6×1/11 =55+6/66/1 5/66 =61/66/66 5/66 =61/66/61/66 =61/66×66/61 =1 ==tanπ/4 ...

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Byju's Answer

Standard XII

Mathematics

Basic Trigonometric Identities

i If tan A =5...

Question

(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 $A - B = \frac{π}{4}$

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Solution

We have
$t a n A = \frac{5}{6}$ and $t a n B = \frac{1}{11}$
Now,
$t a n (A + B) = \frac{t a n A + t a n B}{1 - t a n A . t a n B}$
$= \frac{\frac{5}{6} + \frac{1}{11}}{1 - \frac{5}{6} \times \frac{1}{11}}$
$= \frac{\frac{55 + 6}{66}}{1 - \frac{5}{66}}$
$= \frac{\frac{61}{66}}{\frac{66 - 5}{66}}$
$= \frac{\frac{61}{66}}{\frac{61}{66}}$
$= \frac{61}{66} \times \frac{66}{61}$
=1
= $= t a n \frac{π}{4}$
$\Rightarrow t a n (A + B) = t a n \frac{π}{4}$
$\Rightarrow A + B = \frac{π}{4}$
Hence proved.
(ii) We have,
$t a n A = \frac{m}{m - 1}$ and $t a n B = \frac{m}{2 m - 1}$
Now, $t a n (A - B) = \frac{t a n A - t a n B}{1 + t a n A . t a n B}$
$= \frac{\frac{m}{m - 1} - \frac{1}{2 m - 1}}{1 + \frac{m}{m - 1} \times \frac{1}{2 m - 1}}$
$= \frac{\frac{m (2 m - 1) - (m - 1)}{(m - 1) (2 m - 1)}}{1 + \frac{m}{(m - 1) (2 m - 1)}}$
$= \frac{\frac{m (2 m - 1) - (m - 1)}{(m - 1) (2 m - 1)}}{\frac{(m - 1) (2 m - 1) + (m)}{(m - 1) (2 m - 1)}}$
$= \frac{m (2 m - 1) - (m - 1)}{(m - 1) (2 m - 1) + (m)}$
$= \frac{2 m^{2} - m - m + 1}{2 m^{2} - m - 2 m + 1 + m}$
$= \frac{2 m^{2} - m - m + 1}{2 m^{2} - 2 m + 1}$
$= \frac{2 m^{2} - 2 m + 1}{2 m^{2} - 2 m + 1}$
=1
$∴ t a n (A - B) = 1 = t a n (\frac{π}{4})$
$\Rightarrow t a n (A - B) = t a n (\frac{π}{4})$
$\Rightarrow A - B = (\frac{π}{4})$
Hence proved.

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Similar questions

Q. If

t a n A = \frac{5}{6}

t a n B = \frac{1}{11}

, prove that :

A + B = \frac{π}{4}

Q. If

A + B + C = π

, prove that

tan \frac{A}{2} tan \frac{B}{2} + tan \frac{B}{2} tan \frac{C}{2} + tan \frac{C}{2} tan \frac{A}{2} = 1

Q. (i) If

\tan A = \frac{5}{6} and \tan B = \frac{1}{11}

, prove that

A + B = \frac{π}{4}

.
(ii) If

\tan A = \frac{m}{m - 1} and \tan B = \frac{1}{2 m - 1}

, then prove that

A - B = \frac{π}{4}

Q. If

A = 60^{o}

and

B = 30^{o}

, then prove that

tan (A - B) = \frac{tan A - tan B}{1 + tan A tan B}

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

\frac{1}{a} - \frac{1}{b}

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(i) If tanA=56 and tanB=111, prove that A+B= π4 (ii) If tanA=mm−1 and tanB=m2m−1 then prove that A−B=π4

(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 $A - B = \frac{π}{4}$