Question

10. If the non-zero numbers a,b,c are in AP and
tan-1a., tan-1b, tan-1c
are also in AP, then
(a) a = b = c (b) b²= 2ac (c) a² = bc (d) c² = ab​

Answer 1

SOLUTION

GIVEN

1. The non-zero numbers a,b,c are in AP

$2. \: \: \: \: \sf{ { \tan}^{ - 1}a \: , \: { \tan}^{ - 1} b \: ,\: { \tan}^{ - 1}c \: \: \: are \: in \: AP\: }$

TO CHOOSE THE CORRECT OPTION

Then

(a) a = b = c

(b) b²= 2ac

(c) a² = bc

(d) c² = ab

EVALUATION

Since a, b, c are in AP (Arithmetic progression)

So 2b = a + c .......... (1)

Again

$\sf{ { \tan}^{ - 1}a \: , \: { \tan}^{ - 1} b \: ,\: { \tan}^{ - 1}c \: \: \: are \: in \: AP\: }$

$\implies\sf{2 { \tan}^{ - 1}b \: = \: { \tan}^{ - 1} a + \: { \tan}^{ - 1}c }$

$\implies \displaystyle\sf{ { \tan}^{ - 1} \bigg( \frac{2b}{1 - {b}^{2} } \bigg)\: = \: { \tan}^{ - 1} \bigg( \frac{a + c}{1 - ac} \bigg) }$

$\implies \displaystyle\sf{ \bigg( \frac{2b}{1 - {b}^{2} } \bigg)\: = \bigg( \frac{a + c}{1 - ac} \bigg) }$

$\implies \displaystyle\sf{ \bigg( \frac{2b}{1 - {b}^{2} } \bigg)= \bigg( \frac{2b}{1 - ac} \bigg) } \: ( \because \: 2b = a + c)$

$\implies \displaystyle\sf{1 - {b}^{2} = 1 - ac }$

$\implies \sf{ {b}^{2} = ac }$

$\implies \displaystyle \sf{ { \bigg( \frac{a + c}{2} \bigg)}^{2} = ac }$

$\implies \displaystyle \sf{ { \big( {a + c} \big)}^{2} =4 ac }$

$\implies \displaystyle \sf{ { \big( {a + c} \big)}^{2} - 4 ac = 0 }$

$\implies \displaystyle \sf{ { \big( {a - c} \big)}^{2} = 0 }$

$\implies \displaystyle \sf{a = c} \: \: \: ....(2)$

From Equation (1) we get

$\sf{2b = 2a \: }$

$\implies \displaystyle \sf{ b = a } \: \: ....(3)$

Equation (2) & (3) gives a = b = c

FINAL ANSWER

The correct option is

(a) a = b = c

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LEARN MORE FROM BRAINLY

tan⁻¹(x/y)-tan⁻¹(x-y/x+y)=.......(x/y≥0),

Select Proper option from the given options.

(a) π/4

(b) π/3

(c) π/2

(d) π

https://brainly.in/question/5596504

Answer 2

$\huge{\boxed{\rm{\red{Answer}}}}$

☞ 2b = a + c $\large\purple{\texttt{Equation 1}}$

☞ Now 2tan-¹b = tan-¹a + tan-¹c

☞ or 2b / 1 - b² = a + c / 1 - ac

☞ Put 2b = a + c

☞ b² = ac by $\large\purple{\texttt{Equation 1}}$ that is a , b, c are in G.P

☞ Also , 4b² = 4ac

☞ or ( a + c )² - 4ac = 0 by $\large\purple{\texttt{Equation 1}}$

☞ ( a - c )² = 0

☞ Therefore, a = c = b by $\large\purple{\texttt{Equation 1}}$

@Itzbeautyqueen23

Hope it's helpful

Thank you :]