pls tell me what is the answer
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Step-by-step explanation:
1
Secondary School
Math
6 points
Prove that-(1+tan^2A/1+cot^2A)=(1-tanA/1-cotA)^2=tan^2
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byFunkyishitap6ihwv 02.04.2018
Answers

oswin
Virtuoso
L.H.S
1+tan^2A/1+cot^2A
=(1+sin^2A/cos2A)/(1+cos^2A/sin^2A)
=((cos^2A+sin^2A)/cos^2A)/(sin^2A+cos^2A)/sin^2A
=(1/cos^2A)/(1/sin^2A)
=1/cos^2A*sin^2A/1
=sin^2A/cos^2A
=tan^2A
M.H.S
(1-tanA/1-cotA)^2
=(1+tan^2A-2tanA)/(1+cot^2-2cotA)
=(sec^2A-2tanA)/(cosec^2A-2cosA/sinA)
=(sec^2A-2*sinA/cos A)/(cosec^2A-2cosA/sinA)
=(1/cos^2A-2sinA/cosA)/(1/sin^2A-2cosA/sinA)
=[(1-2sinAcosA)/cos^2A]/[(1-2cosAsinA)sin^2A]
=(1-2sinAcosA)/cos^2A*sin^2A/(1-2sinAcosA)
=sin^2A/cos^2A
=tan^2A
R.H.S
=tan^2A
Answer:
Given:
∠B = 35° and ∠C = 65°
Formula:
Sum of all three angles of a triangle is 180°.
A greater angle of a triangle is opposite a greater side.
Calculation:
∠A + ∠B + ∠C = 180°
⇒ ∠A + ∠35° + ∠65° = 180°
⇒ ∠A + 100° = 180°
⇒ ∠A = 180° – 100°
∴ ∠A = 80°
If AD is bisector of ∠BAC, then
∠BAD = ∠CAD
∠BAD + ∠CAD = 80°
⇒ ∠BAD + ∠BAD = 80°
⇒ 2∠BAD = 80°
⇒ ∠BAD = 40°
∠BAD = ∠CAD = 80°/2 = 40°
In ΔBAD
∠B + ∠BAD + ∠BDA = 180°
⇒ 35° + 40° + ∠BDA = 180°
⇒ 75° + ∠BDA = 180°
⇒ ∠BDA = 180° – 75°
⇒ ∠BDA = 105°
∠BDA + ∠CDA = 180°
⇒ 105 + ∠CDA = 180° [Liner pair
∠CDA = 180° – 105°
⇒ ∠CDA = 75°
In ΔABC
∠A = 80°, ∠B = 35° and ∠C = 65°
BC > AB > AC ----(1)
In ΔABD
∠A = 40°, ∠B = 35° and ∠D = 105°
AB > BD > AD ----(2)
In ΔACD
∠A = 40°, ∠C = 65° and ∠D = 75°
AC > AD > CD ----(3)
From equation (2) and equation (3)
∴ BD > AD > CD