Q.$\cos 2\theta\,\cos 2\phi + \sin^2(\theta - \phi) - \sin^2(\theta + \phi)$ is equal to
(A) $\sin 2(\theta + \phi)$
(B) $\cos 2(\theta + \phi)$
(C) $\sin 2(\theta - \phi)$
(D) $\cos 2(\theta - \phi)$

Double-angle on $80^\circ=2\cdot40^\circ$:

First term:

Second term:

Add: …

The problem asks us to evaluate an expression involving $\sin^2(A+B)$ and $\cos^2(A+B)$, given that $\tan A$ and $\tan B$ are the roots of a quadratic equation. The key idea is to first find $\tan(A+B)$ using the properties of roots of a quadratic equation and the tangent addition formula. Once we have $\tan(A+B)$, we can easily determine $\sin^2(A+B)$ and $\cos^2(A+B)$ using trigonometric identities.

Here's how we approach this problem:

1. Identify the sum and product of roots: For a quadratic equation $ax^2 + bx + c = 0$, if $r_1$ and $r_2$ are its roots, then the sum of roots $r_1 + r_2 = -b/a$ and the product of roots $r_1 r_2 = c/a$. In this problem, the roots are $\tan A$ and $\tan B$.

2. Apply the tangent addition formula: The formula for $\tan(A+B)$ directly relates the sum and product of $\tan A$ and $\tan B$.

   $\tan(A+B) = \frac{\tan A + \tan B}{1 - \tan A \tan B}$

3. Convert $\tan(A+B)$ to $\sin^2(A+B)$ and $\cos^2(A+B)$: Once we have the value of $\tan(A+B)$, we can use fundamental trigonometric identities to find $\sin^2(A+B)$ and $\cos^2(A+B)$. A common way is to use $\sec^2 \theta = 1 + \tan^2 \theta$, from which $\cos^2 \theta = 1/\sec^2 \theta$, and then $\sin^2 \theta = 1 - \cos^2 \theta$.

Let's work through the steps:

1. Extract information from the quadratic equation:

   The given quadratic equation is $2x^2 - 9x - 16 = 0$.

   The roots are $\tan A$ and $\tan B$.

   Using Vieta's formulas:

   Sum of roots: $\tan A + \tan B = -\frac{(-9)}{2} = \frac{9}{2}$.

   Product of roots: $\tan A \tan B = \frac{-16}{2} = -8$.

2. Calculate $\tan(A+B)$:

   Now, we use the tangent addition formula:

Substitute the values we found:

3. Find $\sin^2(A+B)$ and $\cos^2(A+B)$: Let $\theta = A+B$. We have $\tan \theta = \frac{1}{2}$. We know the identity $\sec^2 \theta = 1 + \tan^2 \theta$. $\sec^2(A+B) = 1 + \left(\frac{1}{2}\right)^2 = 1 + \frac{1}{4} = \frac{5}{4}$ …

Solution

Write both terms over $\sinh$ and $\cosh$ and combine:

Factor the numerator as a difference of squares:

using $\cosh^{2}x-\sinh^{2}x=1$ and $\cosh^{2}x+\sinh^{2}x=\cosh2x$.

For the denominator, $\sinh2x=2\sinh x\cosh x$ gives $\sinh^{2}x\cosh^{2}x=\dfrac{\sinh^{2}2x}{4}$. Therefore …

We start by noticing that the right-hand side of the given equation involves standard angles:

Thus the right-hand side becomes:

So the equation simplifies to:

1. Use the identity linking $\sin^{-1}$ and $\cos^{-1}$. Recall that for any argument in the appropriate domain,

Here we have $\cos^{-1}(2x)$, so we can rewrite $\sin^{-1} x$ in terms of $\cos^{-1} x$ if needed. But more directly, let’s isolate one inverse function.

From $\sin^{-1} x - \cos^{-1}(2x) = \frac{\pi}{6}$, we can write:

2. Take sine of both sides. Taking $\sin$ of both sides (valid because both sides lie in a range where sine is one-to-one) gives:

Use the sine addition formula:

Here $A = \frac{\pi}{6}$, $B = \cos^{-1}(2x)$. So:

3. Simplify the trigonometric expressions. We know $\sin\frac{\pi}{6} = \frac12$, $\cos\frac{\pi}{6} = \frac{\sqrt{3}}{2}$, and $\cos(\cos^{-1}(2x)) = 2x$. For $\sin(\cos^{-1}(2x))$, recall that if $\theta = \cos^{-1}(2x)$, then $\cos\theta = 2x$ and $\sin\theta = \sqrt{1 - (2x)^2}$ (taking the positive root because $\cos^{-1}$ outputs angles in $[0,\pi]$, where sine is nonnegative). So:

Substituting:

This simplifies to:

4. Solve for $x$. Subtract $x$ from both sides:

Since $\frac{\sqrt{3}}{2} \neq 0$, we must have:

So $x = \frac12$ or $x = -\frac12$.

5. Check domain restrictions. For $\sin^{-1} x$ to be defined, we need $|x| \le 1$ — both values satisfy this. For $\cos^{-1}(2x)$ to be defined, we need $|2x| \le 1$, i.e., $|x| \le \frac12$ — again both satisfy. …

The key insight is to treat each pair $(\cos\theta,\sin\theta)$ as a point on the unit circle, or equivalently as the complex number $e^{i\theta}$. When three such vectors sum to zero, they must be the vertices of an equilateral triangle centered at the origin. That geometric fact unlocks all the algebra.

Let’s work through it cleanly.

1. Rewrite the conditions as a single complex equation. Let $z_1 = e^{i\alpha}$, $z_2 = e^{i\beta}$, $z_3 = e^{i\gamma}$. Then

So $z_1+z_2+z_3 = 0$.

2. What does $z_1+z_2+z_3=0$ imply geometrically? Each $z_k$ lies on the unit circle $|z_k|=1$. Their sum is zero, so they are the vertices of an equilateral triangle centered at the origin. Consequently, they are spaced by $120^\circ$:

where $\omega = e^{2\pi i/3}$ is a primitive cube root of unity, satisfying $1+\omega+\omega^2=0$ and $\omega^3=1$.

3. Now compute the required sum. We want $\cos2\alpha+\cos2\beta+\cos2\gamma$. In complex form,

So

4. Square the sum condition. From $z_1+z_2+z_3=0$, square both sides:

So

5. Evaluate the product sum. Using $z_2 = z_1\omega$, $z_3 = z_1\omega^2$:

Hence

Given: $a=7,\; c=11,\; \cos A=\dfrac{17}{22},\; \cos C=\dfrac{1}{14}$.

Find $b$ (law of cosines with the $\cos C$ data):

(The $\cos A$ data confirms this: $49=b^2+121-2b(11)\tfrac{17}{22}=b^2-17b+121\Rightarrow b^2-17b+72=0\Rightarrow b=9$.)

Apply the tangent identity. In any triangle,

Therefore …

When an equation relates $x$ and $y$ but it's difficult or impossible to express $y$ explicitly as a function of $x$ (i.e., $y = f(x)$), we use a technique called implicit differentiation. The core idea is to differentiate both sides of the equation with respect to $x$, remembering that $y$ is a function of $x$. This means that whenever we differentiate a term involving $y$, we must apply the chain rule, multiplying by $\frac{dy}{dx}$.

Let's apply this to the given equation.

1. Differentiate both sides with respect to $x$: The given equation is $x \cos (k+y) = \cos y$. We differentiate both sides with respect to $x$:

For the left side, we use the product rule $\frac{d}{dx}(uv) = u'v + uv'$ where $u=x$ and $v=\cos(k+y)$.
For the right side, we use the chain rule $\frac{d}{dx}(\cos y) = -\sin y \cdot \frac{dy}{dx}$.

Since $k$ is a constant, $\frac{d}{dx}(k+y) = 0 + \frac{dy}{dx} = \frac{dy}{dx}$.
Substituting this, we get:

2. Rearrange to isolate $\frac{dy}{dx}$: Our goal is to solve for $\frac{dy}{dx}$. We gather all terms containing $\frac{dy}{dx}$ on one side and the other terms on the opposite side:

Factor out $\frac{dy}{dx}$ from the terms on the right side:

Now, divide to solve for $\frac{dy}{dx}$:

3. Substitute $x$ in terms of $y$ to simplify: From the original equation, we know that $x = \frac{\cos y}{\cos(k+y)}$. Substitute this expression for $x$ into the denominator of our $\frac{dy}{dx}$ expression:

To simplify the denominator, find a common denominator:

Now, multiply the numerator by the reciprocal of the denominator:

$$ \frac{dy}{dx} = \frac{\cos^2(k+y)}{\sin(k+y)\cos y - \cos(k+y)\sin y} $$ …

The core idea here is parametric differentiation. When $x$ and $y$ are both given in terms of a third variable (here $t$), you cannot directly differentiate $y$ with respect to $x$. Instead, you find the derivatives of each with respect to $t$, and then take their ratio: $\frac{dx}{dy} = \frac{dx/dt}{dy/dt}$. This works because the $dt$ cancels, just like in the chain rule.

Let’s apply this step by step.

1. Differentiate $x$ with respect to $t$. $x = 5(1 - \sin t)$ The derivative of $1$ is $0$, and the derivative of $-\sin t$ is $-\cos t$. So:

2. Differentiate $y$ with respect to $t$. $y = 5(t + \cos t)$ The derivative of $t$ is $1$, and the derivative of $\cos t$ is $-\sin t$. So:

3. Form the ratio $\frac{dx}{dy}$.

4. Simplify the expression. Notice that $-\cos t$ in the numerator and $1 - \sin t$ in the denominator. Multiply numerator and denominator by $-1$ to get a cleaner form:

This is exactly option (B). But wait — check the sign carefully. Our simplified result is $\frac{\cos t}{\sin t - 1}$, which is not the same as $\frac{\sin t - 1}{\cos t}$ (option A). Let’s verify which one is correct.

5. Check if further simplification matches any option.

   Option (A) is $\frac{\sin t - 1}{\cos t}$, which is the negative reciprocal of our result. Option (B) is $\frac{\cos t}{\sin t - 1}$, which is exactly what we have. Options (C) and (D) are trigonometric forms that might simplify to one of these, but let’s test with a specific $t$ value to be sure.

   Take $t = \frac{\pi}{2}$: …

The key here is to see that the expression mixes side lengths and angle differences. In any triangle, sides are proportional to the sines of opposite angles, so we can rewrite everything in terms of angles alone. That makes the trigonometric simplifications natural.

1. Express sides in terms of sines. By the law of sines, $\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = 2R$, where $R$ is the circumradius. So

2. Rewrite the denominators.

3. Use sum-to-product identities.

Since $A+B+C = \pi$, we have $\frac{B+C}{2} = \frac{\pi - A}{2} = \frac{\pi}{2} - \frac{A}{2}$. Therefore

Substituting: