show that (1+ tan20) (1+tan25)-2
{use tan(A+B)=tan A+ tan B/1-tan A + tan B}
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(1+tan 20)(1+tan 25)-2
=1 + tan 25 + tan 20 + tan20 × tan25 - 2
=tan 20 + tan 25 + tan20 × tan 25 -1
using tan(A+B)=(tan A + tan B)/(1-tan A×tan B)
⇒tan(A+B)×(1-tan A×tan B)=tan A+ tan B
take A=20, B=25
⇒tan(20+25)×(1-tan 20×tan 25)=tan 20 + tan 25
⇒tan(45)×(1-tan 20×tan 25)=tan 20 + tan 25
⇒1×(1-tan 20×tan 25)=tan 20 + tan 25
⇒1 - tan 20×tan 25=tan 20 + tan 25
⇒tan 20 + tan 25 + tan 20 × tan 25 - 1 = 0
So the required answer is 0.
=1 + tan 25 + tan 20 + tan20 × tan25 - 2
=tan 20 + tan 25 + tan20 × tan 25 -1
using tan(A+B)=(tan A + tan B)/(1-tan A×tan B)
⇒tan(A+B)×(1-tan A×tan B)=tan A+ tan B
take A=20, B=25
⇒tan(20+25)×(1-tan 20×tan 25)=tan 20 + tan 25
⇒tan(45)×(1-tan 20×tan 25)=tan 20 + tan 25
⇒1×(1-tan 20×tan 25)=tan 20 + tan 25
⇒1 - tan 20×tan 25=tan 20 + tan 25
⇒tan 20 + tan 25 + tan 20 × tan 25 - 1 = 0
So the required answer is 0.
explain the second step
just expand.(a+b)(c+d)=ac+ad+bc+bd
SIR IT IS 1- tan A
(1-tanA+tanB)
see the formula in your book...
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