Question

tan⁻¹ 15 +tan⁻¹ 17 +tan⁻¹ 13 + tan⁻¹ 18 = π4 सिद्ध कीजिए

Answer 1

Step-by-step explanation:

LHS = tan^-1(1/5) + tan^-1(1/7) + tan^-1(1/3) + tan^-1(1/8)

we know, tan^-1x + tan^-1y = tan^-1(x + y)/(1 - xy)

where , xy < 1

= [tan^-1(1/5) + tan^-1(1/7) ] + [tan^-1(1/3) + tan^-1(1/8) ]

= tan^-1{(1/5 + 1/7)/(1 - 1/5 × 1/7)} + tan^-1{(1/3 + 1/8)/(1 - 1/3 × 1/8)}

= tan^-1{(12/35}/(34/35)} + tan^-1{(11/24)/(23/24}

= tan^-1{6/17} + tan^-1{11/23}

= tan^-1{(6 × 23 + 17 × 11)/17 × 23}/{1 - 6/17 × 11/23}

= tan^-1{(138 + 187)/(391 - 66)}

= tan^-1{325/325}

= tan^-1(1) = π/4 = RHS

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Answer 2

Given   :  tan⁻¹ 1/5 +tan⁻¹ 1/7 +tan⁻¹ 1/3 + tan⁻¹ 1/8 = π4

To find :   सिद्ध कीजिए

Solution:

tan⁻¹ 1/5 +tan⁻¹ 1/7 +tan⁻¹ 1/3 + tan⁻¹ 1/8 = π4

दोनों तरफ Tan लेने पर

=> tan ( tan⁻¹ 1/5 +tan⁻¹ 1/7 +tan⁻¹ 1/3 + tan⁻¹ 1/8)  = Tan π4

LHS = tan ( (tan⁻¹ 1/5 +tan⁻¹ 1/7) +(tan⁻¹ 1/3 + tan⁻¹ 1/8))

माना  A =  tan⁻¹ 1/5 +tan⁻¹ 1/7   B = tan⁻¹ 1/3 + tan⁻¹ 1/8

tan(A + B) = (tanA + tanB) / (1 - tanAtanB)

A =  tan⁻¹ 1/5 +tan⁻¹ 1/7

Tan A = Tan(tan⁻¹ 1/5 +tan⁻¹ 1/7  )

= (1/5  + 1/7 )/( 1-  (1/5)(1/7))

= (7 + 5)/(35 - 1)

= 12/34

= 6/17

=> A = tan⁻¹ 6/17

B +

tanB = tan(tan⁻¹ 1/3 + tan⁻¹ 1/8)

= ( 1/3 + 1/8)/(1 - (1/3)(1/8))

= (8 + 3)/ (24 - 1)

= 11/23

=> B = tan⁻¹ 11/23

LHS = (6/17 + 11/23)/(1 - (6/17)(11/23)

= (138 + 187)/(391 - 66)

= 325/325

= 1

RHS = Tan π4

= 1

LHS = RHS

=> tan⁻¹ 1/5 +tan⁻¹ 1/7 +tan⁻¹ 1/3 + tan⁻¹ 1/8 = π4

QED

इति सिद्धम

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