IIT JEE , AIEEE Forum - Mathematics , Trignometry

aks_0123456

Q1 If x and y be solution of @ for equation atan@ +bsec@ =c then prove tan(x+y)=2ac/(a² - c²)

Q2 find sin2@ if sin(@/2) +cos(@/2) = -1/2

Q3 prove that triangle ABC is equilateral iff tanA + tanB + tanC =3^{[ ]}

3

Q4 if sinx +sin²x =1 find value of cos¹²x +3cos¹⁰x +3cos⁸x +cos⁶x - 1

Plzzzzzz provide full solution

prash_shan_jpr

In 2 ques. square the quation then sin^2 A + cos ^2 A = 1

u have sin A

find cos A

use 2 sin A cos A

prash_shan_jpr

In the que

sinx + sin ^2 x = 1

use sin ^2 x = 1 - cos ^2 x

take it to the lhs

u will have sinx=cos ^2 x

put it in the gven equation

solve it

prash_shan_jpr

Be the 1st to rate me pls..!!

punnima

in q2) sin

/2 + cos

/2 = - 1

2

on squaring both sides

sin ²

/2+ cos²

/2 + 2 sin

/2 cos

/2 = 1

4

1+ sin2

=1/4

sin 2

=(1/4)-1= -3/4

punnima

If x and y be solution of @ for equation atan@ +bsec@ =c then prove tan(x+y)=2ac/(a² - c²)

write sec@ as tan²@ +1
a (1+tan²@) + btan@ =c

asec²@ +bsec@ -a-c=0

tan@ = -b +

(b²-4a(-a-c))

2a

= -b +

(b²+4a²+4ac)

2a

therfore x= tan^-1[ -b+

(b^2+4a^2+4ac) ]

2a

and y= tan^-1[ -b-

(b^2+4a^2+4ac) ]

2a

now, tan(x+y)= tanx + tan y

1- tanx tan y

= -2b/2a + 0

1-( b^2 - (b^2+4a^2+4ac) ]

4a^2 4a^2

punnima

is my soln so far rite???
i think after this d ans shud b rite..............
but i donn seem 2 get it

raulrag009

Q1

Here

atan@ + bsec@ =c

bsec@ = c-atan@

squaring both sides

b²(1+tan²@)=c²+a²tan²@-2actan@

(b²-a²)tan²@ +2actan@+b²-c²=0

since x,y are the solutions for @

Let tanx and tany be the roots of the above eqn

sum of roots = -B/A

product of roots =C/A

hence

tanx+tany = -2ac/(b²-a²)

and

tanxtany = (b²-c²)/(b²-a²)

Now tan(x+y)= (tanx+tany)/(1-tanxtany)

= -2ac/(b²-c²) / 1 - (b²-c²)/(b²-a²)

tan(x+y) =2ac/(a²-c²)

raulrag009

Q3

As A+B+C=

A+C=

-B

taking tan on both sides

tan(A+C)= -tanB

(tanA+tanC)/(1-tanAtanC) = -tanB

tanA+tanC= -tanB+tanAtanBtanC

tanA+tanB+tanC = tanAtanBtanC

thus

tanAtanBtanC= 3

3

From this

tanA=

3

tanB=

3

tanC=

3

Thus A=B=C=Pie/3

punnima

gud!
thanx 4 providing a better soln

hsbhatt

sinx+sin²x = 1

Hence sinx = cos²x

The given expression becomes

sin⁶x + 3sin⁵x + 3sin⁴x+sin³x-1

Now, sinx+sin²x-1 = 0

We know that if a+b+c = 0, then a³+b³+c³ = 3abc

Hence sin³x+sin⁶x-1 = -3sin³x

Hence

sin⁶x + 3sin⁵x + 3sin⁴x+sin³x-1 = 3sin⁵x + 3sin⁴x +(sin³x+sin⁶x-1)

= 3sin⁵x + 3sin⁴x -3sin³x = 3sin³x(sin²x+sinx-1) = 0

hsbhatt

Q3 has not been solved convincingly.

When the question says iff p then q rather than just if p then q, you have to prove p

q and q

p. In this case of course neither has been proved.

Proving that if ABC is equilateral tanA+tanB+tanC = 3

3 is trivial.

Now, we have to prove that if for a triangle ABC, tanA+tanB+Tanc = 3

3 then A=B=C = 60^o.

Here you must invoke the inequality, that in a triangle

tanA+tanB+tanc

3

3 with equality occuring only in the case where A = B = C.

Since tanA+tanB+Tanc = 3

3 this must mean A=B=C = 60^o.

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