Q.If $m\sin\theta = n\sin(\theta + 2\alpha)$, then prove that $\tan(\theta + \alpha)\cot\alpha = \dfrac{m + n}{m - n}$.

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

First term:

Second term:

Add:

The stated $\theta=\tfrac{11\pi}{7}$ does not appear in the expression and does not affect the value.

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 $\sec^2 \theta = 1 + \tan^2 \theta$.

Since $\cos^2 \theta = \frac{1}{\sec^2 \theta}$:

Now, use the identity $\sin^2 \theta = 1 - \cos^2 \theta$:

> [!TIP]
> Alternatively, you can visualize a right-angled triangle where $\tan \theta = \frac{\text{opposite}}{\text{adjacent}} = \frac{1}{2}$. The hypotenuse would be $\sqrt{1^2 + 2^2} = \sqrt{5}$.
> Then $\sin \theta = \frac{1}{\sqrt{5}}$ and $\cos \theta = \frac{2}{\sqrt{5}}$.
> So, $\sin^2 \theta = \left(\frac{1}{\sqrt{5}}\right)^2 = \frac{1}{5}$ and $\cos^2 \theta = \left(\frac{2}{\sqrt{5}}\right)^2 = \frac{4}{5}$.

4. Evaluate the expression:

The expression we need to evaluate is $9\sin^2(A+B) - \cos^2(A+B)$.

Substitute the values we found for $\sin^2(A+B)$ and $\cos^2(A+B)$:

The value of the expression is $1$.

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

Check ($x=1$): LHS $=\coth^{2}1-\tanh^{2}1\approx1.724-0.580=1.144$; RHS $=4\cosh2\,(\operatorname{csch}2)^{2}\approx1.144$. $\checkmark$

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. But we must also check the original equation: if $x = -\frac12$, then $\sin^{-1}(-\frac12) = -\frac{\pi}{6}$ and $\cos^{-1}(-1) = \pi$. Then left side is $-\frac{\pi}{6} - \pi = -\frac{7\pi}{6}$, not equal to $\frac{\pi}{6}$. So $x = -\frac12$ is extraneous (likely introduced by squaring or the sine step). Hence $x = \frac12$ is the only valid solution.

6. Evaluate the target expression. We need:

Substitute $x = \frac12$:

7. Use the tangent addition formula. Recall:

provided the sum is in the principal range. Here $a = \frac12$, $b = \frac13$:

So:

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

Therefore $z_1^2+z_2^2+z_3^2 = 0$, and taking the real part gives

6. Match with the options. Option (D) is $\cos(\alpha+\beta)+\cos(\beta+\gamma)+\cos(\gamma+\alpha)$. Using $e^{i(\alpha+\beta)} = z_1z_2$, etc., we have

So option (D) also equals zero, matching our result.

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

Substitute: $c-a=11-7=4,\; c+a=18$, so

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 $\frac{dy}{dx}$.