# prove that- (1) (tanθ)/(1-cotθ)+(cotθ)/(1-tanθ)=1+tanθ+cot θ (2) (tan^3A-1)/(tanA-1)=sec^2A+tanA

hala718 | High School Teacher | (Level 1) Educator Emeritus

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1) (tano) /(1-coto) + coto/(1-tan o)= 1+tano+coto

we know that tan= sin/cos and cot=cos/sin

Now by substituting:

(sino/coso)/(1-coso/sino) + (coso/sino)/(1-sino/coso)

= (sin^2)/cos(sin-cos) + cos^2(sin(cos-sin)

= (sin^2)/(cos(sin-cos) - cos^2/sin(sin-cos)

= (sin)(sin)/cos(sin-cos) - (cos)(cos)/sin(sin-cos)

= (sin/cos)*(sin/(sin-cos) - (cos)/(sin)*( cos/(sin-cos)

= tan*sin(1/(sin-cos) - cot*cos(1/(sin-cos)

factor 1/(sin-cos)

= (1/(sin-cos) [tan*sin -cot*cos)

=(1/(sin-cos)[(sin^3-cos^3)/(sin*cos)]

= (1/(sin-cos) * (sin-cos)(sin^2+sin*cos +cos^2)/(sin*cos)

= (sin^2+sin*cos+cos^2)/sin*cos

= sin/cos + 1 + cos/sin = tan + 1 +cot = tan+cot +1

2)(tan^3(A)-1)/(tanA -1) =sec^2 A + tanA

tan= sin/cos

==> [(sin^3(a)/cos^3(a)) -1 ]/ [(sin/cos) -1)]

= [(sin^3 -cos^3)/(cos^3)]/[(sin-cos)/cos]

= (sin-cos)(sin^2-sincos+cos^2)cos / (sin-cos)cos^3

= (sin^2 -sin*cos +cos^2) /cos^2

= (sin^2/cos^2 + sin*cos/cos^2+ cos^2/cos^2

= (tan^2) +(tan)+ 1

= sin^2/cos^2 + 1 + tan

= (sin^2+cos^2)/cos^2 +tan

= 1/cos^2 + tan

= sec^2 + tan

lochana2500 | Student, Undergraduate | (Level 1) Valedictorian

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1). L:H:S:≡ tanθ/(1-cotθ) + cotθ/(1-tanθ)

= tanθ /[(tanθ-1)/tanθ] + (1/tanθ)/ (1-tanθ)

= tan²θ/(tanθ-1) + 1/tanθ(1-tanθ)

=tan²θ/(tanθ-1) - 1/tanθ(tanθ-1)

=[1/(tanθ-1)] * [tan²θ - 1/tanθ]

=[1/(tanθ-1)] * [(tan³ - 1)/tanθ]

⇒ x³ - y³ = (x - y)(x² + y² + xy)

=[1/(tanθ-1)] * [(tanθ-1)(tan²θ + tanθ +1)/tanθ]

= (tan²θ + tanθ +1)/ tanθ

= tanθ + cotθ + 1

= R:H:S

2). L:H:S ≡ (tan³A-1)/(tanA-1)

⇒ x³ - y³ = (x - y)(x² + y² + xy)

=(tanA-1)(tan²A + tanA + 1)/(tanA-1)

= tan²A + tanA + 1

⇒ 1 + tan²A = sec²A

= sec²A + tanA

= R:H:S

jeyaram | Student, Undergraduate | (Level 1) Valedictorian

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(1) (tanθ)/(1-cotθ)+(cotθ)/(1-tanθ)=1+tanθ+cot θ

L.H.S=(tanθ)/(1-cotθ)+(cotθ)/(1-tanθ)

=(sinθ/cosθ)/{1-(cosθ/sinθ)}+(cosθ/sinθ)/{1-(sinθ/cosθ)}

=(sinθ/cosθ)/{(sinθ-cosθ)/sinθ}+(cosθ/sinθ)/{(cosθ-sinθ)/cosθ}

=sin^2 θ/(cosθ[sinθ-cosθ]) + cos^2 θ/(sinθ[cosθ-sinθ])

=sin^2 θ/(cosθ[sinθ-cosθ]) - cos^2 θ/(sinθ[sinθ-cosθ])

=1/(sinθ-cosθ){sin^2 θ/cosθ - cos^2 θ/sinθ}

=1/(sinθ-cosθ){(sin^3 θ - cos^3 θ)/sinθcosθ}

=1/(sinθ-cosθ){[(sinθ - cosθ)(sin^2 θ+sinθcosθ+cos^2 θ)]/sinθcosθ}

=(sin^2 θ+sinθcosθ+cos^2 θ)/sinθcosθ

=sin^2 θ/sinθcosθ + sinθcosθ/sinθcosθ + cos^2 θ/sinθcosθ

=sinθ/cosθ + 1 + cosθ/sinθ

=tanθ+1+cotθ

=1+tanθ+cot θ

=R.H.S

(2) (tan^3 A-1)/(tanA-1)=sec^2 A+tanA

L.H.S=(tan^3 A-1)/(tanA-1)

=([sin^3 A/cos^3 A]-1)/([sinA/cosA]-1)

=(sin^3 A-cos^3 A)/(cos^2 A[sinA-cosA])

={(sinA - cosA)(sin^2 A+sinAcosA+cos^2 A)}/(cos^2 A[sinA-cosA])

=(sin^2 A+sinAcosA+cos^2 A)/cos^2 A

=(sin^2 A+cos^2 A+ sinAcosA)/cos^2 A

=(1+sinAcosA)/cos^2 A     [because sin^2 A+cos^2 A=1]

=1/cos^2 A + sinAcosA/cos^2 A

=1/cos^2 A + sinA/cosA

=sec^2 A+tanA

=R.H.S