sin a+cos a+tan a+sec a+cosec a+cot a=7.then prove that sin 2a is the root of x^2-44x+36

Question

I am not havin g any idea to solve it.

Answer 1

Using standard trig identities, and multiplying through by sin a cos a, the starting equation can be

rewritten as

 

 (sin a  +  cos a) (1 + sin a  cos a ) =  7 sin a  cos a   - 1

 

Next, we replace  sin a cos a  by   (1/2) sin 2a

 

Next use the half angle formula to replace sin a + cos a with

 sqrt [ (1 - cos 2a)/2 ]  + sqrt [ (1+ cos 2a)/2 ]

 

Next square both sides.  The  result of squaring the sum of the square roots simplifies to 1 + sin 2a.

 

The result, after some rearranging, is a quadratic equation in sin 2a.   If we substitute x = sin 2 a,

the equation can be written as x² - 44 x + 36 =0

 

Answer 2

Here is a hint:

 

let us rewrite the stated equality as sinA+cosA+1/(sinAcosA)+(sinA+cosA)/(sinAcosA)=7...here, we have just combined secA and cscA, along with tanA and cotA via trig identities.

 

Know that if sin2A is a root of the quadratic function x^2-44x+36, then sin2A is the smaller root (since the larger root is outside the range of the sine function considering we are not considering complex extensions here).

 

Define z:=sinA+cosA and use z instead in the above expression (substitution) completely so that there is no "A" in the equality (hint: think about what happens when you square z).

 

The above substitution should generate a quadratic with two roots (only one of which is valid due to range considerations), which you can then reconcile with the root obtained from x^2 -44x+36.

 

Please comment if you still have doubts...but I think this should provide a good jumping off point.

 

Good luck:)