Ed's trigonometric expression

Since $\tan^{2} \theta + 1 = \sec^{2} \theta$ , the equation above will be $\sin^{3} \theta \times \cos^{3} \theta$ which can be reduced by the double angle identity of sine

$\sin 2\theta = 2\sin \theta \cos \theta$

$\frac{\sin 2\theta}{2} = \sin \theta \cos \theta$

so, let $u = \frac{\sin 2\theta}{2}$

$u^{3} = (\frac{\sin 2\theta}{2})^{3}$ since the range of $\sin \theta$ is $-1 \leq y \leq 1$ , with $1$ as the maximum value, substituting it to $\sin 2\theta$ should give the maximum value of the expression. Thus,

$(\frac{1}{2})^{3} = \frac{1}{8}$

so, $1 + 8 = 9$

$\tan^2 \theta +1 = \sec^2 \theta = \frac{1}{\cos^2 \theta}$ , so the expression is equal to $\sin^3 \theta \times \cos^3 \theta = (\sin^2 \theta \times \cos^2 \theta)^\frac{3}{2} \leq ((\frac{\sin^2 \theta + \cos^2 \theta}{2})^2)^\frac{3}{2}=\frac{1}{8}$ , by the AM-GM inequality, so the answer is $9$ , as the equality is attained when $\sin \theta = \cos \theta$

$\frac{sin^{3}(\theta)cos(\theta)}{tan^{2}(\theta)+1}$

= $\frac{sin^{3}(\theta)cos(\theta)}{ \frac{sin^{2}(\theta)}{cos^{2}(\theta)} + \frac{cos^{2}(\theta)}{cos^{2}(\theta)} }$

= $\frac{sin^{3}(\theta)cos(\theta)}{ (\frac{sin^{2}(\theta)+cos^{2}(\theta)}{cos^{2}(\theta)}) }$

= $\frac{sin^{3}(\theta)cos(\theta)}{ (\frac{1} {cos^{2}(\theta)}) }$

= $sin^{3}(\theta)cos^{3}(\theta)$

= $\frac{\sin^{3}(2\theta)}{8}$

Since $\sin(x)$ is bounded from -1 to 1, the maximum value of $\sin(x)$ is 1. Hence the maximum value of $\sin(2\theta)$ is 1. So our maximum value becomes

$\frac{1^{3}}{8} = \frac{1}{8}$

I'm plot in Graphmatica too

Let theta = x We can simplify this expression to sin^3x*cos^3x To maximise this we have to make cos x =sin x cos x =sinx=1/(2^0.5) by solving we get answer as 9

(tanθ^2)+1=[(sin^2)+(cosθ^2)]/cosθ^2 Numerator becomes (sinθcosθ)^3 sinθcosθ=sin2θ/2 Thus (sinθcosθ)^3=(sin2θ/2)^3 The maximum value of sin function is 1 Thus value is (1/2)^3 As required a+b = 1+8 = 9

On solving above expression we can see that expression becomes

{sin(x)cos(x)}^3 ...........(i)

Now by multiplying 8 in numerator and denominator in exp.(i) it becomes

[{2sin(x)cos(x)}^3]/8

= [{sin(2x)}^3]/8. ............(ii) {sin(2x) = 2sin(x)cos(x)}

And we know that value of sin(2x) lies in [0,1].

Therefore, maximum value of exp. (ii) is 1/8. As we can see clearly that a = 1, b = 8 and a + b = 9.

1+tan^2 (x) = sec^2 (x)...the question simplifies to sin^3 (x) + cos^3 (x)...Now the maximum value will be attained only when sin x = cos x ...(if we do by Trial and Error..we note that whenever the angle(x) increases sin(x) increases and cos(x) decreases so x cannot be equal to 0,30,60,90... (we wont go any further as cos value becomes negative after 90 degrees hence we wont get any POSITIVE(and/or) coprime value.so only value of x is 45 degrees where both sin x = cos x..hence we plug in the values to get the result...therefore to sin^3 (45) + cos^3 (45)=1/8...(a/b) therefore a+b=1+8=9

1+( $\tan \theta$ ^2)= $\sec \theta$ ^2

1/ $\sec \theta$ ^2=( $\cos \theta$ ^2)

Thus numerator becomes [ $\sin \theta$ $\cos \theta$ ]^3

Which is [ $\sin2 \theta$ /2]^3

maximum value of sin function is 1

thus answer is 1/8=a/b

a+b=9

Instantly, we noticed that

$\tan^2\theta+1=\frac{1}{\cos\theta}$

So, it becomes

$\sin^3\theta\cos^3\theta$

$=(\sin\theta\cos\theta)^3$

$=\Big(\frac{1}{2}(\sin2\theta+\sin0)\Big)^3$ Note:The $\text{maximum value}$ of $\sin2\theta=1$

$=(\frac{1}{2})^3$

$=\frac{1}{8}$

Hence, tha value of $a+b=9$

Get sin^2θ+cos^2θ=1 and divide for cos^2, you conclued that tan^2θ+1=sec^θ, soon, the equation sougth is the maximum value of sin^3.cos^3θ. Remember that sin2θ=2.sinθ.cosθ, ( )^3. Maximum value for sinθ is 1, then, sin^3θ is 1 too. 8.sin^3θ.cos^3θ=1 => sin^3θ.cos^3θ=1/8

(tanθ)^2+1=secθ so equation become (sinθ)^3(cosθ)^3=1/8(sin2θ)^3 since the maximum of sin2θ is 1 ,so maximum of (sinθ)^3(cosθ)^3 is 1/8

The expression is the same sin3(x) * cos3(x) is equal to (sin(x) * cos (x) ) ^3 = ((1/2)sin(2X))^3 =(1/8)* sin(2x)^3 thenthe maximun values is 1/8 with sin(x) = 1 Later a+b = 9 =D

Since tan2(x) + 1 = 1/cos2(x)

Equation results in f(x) = sen3(x)cos3(x)

diferentiating and equaling to 0 we get the maximum point (pi/4), replacing in the original equation we obtain 1/8 so 1+8=9

1 + tan^2 phi = 1/ cos^2 phi => A = sin^3 phi. cos^3 phi = 1/8 sin^3 2phi <= 1/8

f(theta) = sin^3(theta) cos^3(theta), setting sin(theta)^2 = t we have f(t) = t^(3/2) (1-t)^(3/2). By the symmetry of f(t) wrt t = 1/2, the maximum is either at t = 1/2 or t = 0 or t = 1, which gives max(f(t)) = f(1/2) = 1/8

we have that tg x squared + 1 = sec x squared. So we are going to have the expression Sen^3x . Cos^3x and the maximum value of the expression is when the sen x = cos x.. 1/8 = 9

We know that tan^2θ+1=sec^2θ, the equation above will be sin^3θ×cos^3θ which can be reduced by the double angle identity of sine

sin2θ=2sinθcosθ sin2θ/2=sinθcosθ so, let x=sin2θ/2 x^3=(sin2θ/2)^3 since the range of sinθ is −1≤y≤1, with 1 as the maximum value, substituting it to sin2θ should give the maximum value of the expression. Thus,

(1/2)^3=1/8 so, 1+8=9

The denominator which is $\tan^2{\theta}$ $