Solve the following. \({\left( {\frac{{sin27^\circ }}{{cos63^\circ }} - \frac{{cos27^\circ }}{{sin63^\circ }}} \right)^2}\)
0
The problem asks us to evaluate the expression \({\left( {\frac{{sin27^\circ }}{{cos63^\circ }} - \frac{{cos27^\circ }}{{sin63^\circ }}} \right)^2}\). This expression involves trigonometric ratios of angles that seem different (27° and 63°), but they are related through the concept of complementary angles.
Two angles are called complementary if their sum is 90°. In this case, 27° + 63° = 90°, so 27° and 63° are complementary angles.
We can use the following complementary angle identities:
Using these identities, we can relate the trigonometric ratios of 63° to those of 27°:
Now, substitute these equivalent expressions back into the original equation:
\({\left( {\frac{{sin27^\circ }}{{cos63^\circ }} - \frac{{cos27^\circ }}{{sin63^\circ }}} \right)^2}\)
Substitute \(\cos63^\circ = \sin27^\circ\) and \(\sin63^\circ = \cos27^\circ\):
\({\left( {\frac{{sin27^\circ }}{{sin27^\circ }} - \frac{{cos27^\circ }}{{cos27^\circ }}} \right)^2}\)
Assuming \(\sin27^\circ \ne 0\) and \(\cos27^\circ \ne 0\) (which is true for 27°), the fractions simplify:
So the expression becomes:
\({\left( {1 - 1} \right)^2}\)
Perform the subtraction inside the parenthesis:
\({\left( 0 \right)^2}\)
Finally, calculate the square:
\({0^2} = 0\)
Let's break down the solution process:
The final value of the given expression is 0.
| Expression Part | Transformation/Identity | Result |
|---|---|---|
| \(\cos63^\circ\) | \(\cos(90^\circ - 27^\circ)\) | \(\sin27^\circ\) |
| \(\sin63^\circ\) | \(\sin(90^\circ - 27^\circ)\) | \(\cos27^\circ\) |
| \(\frac{{sin27^\circ }}{{cos63^\circ }}\) | \(\frac{{sin27^\circ }}{{sin27^\circ }}\) | 1 |
| \(\frac{{cos27^\circ }}{{sin63^\circ }}\) | \(\frac{{cos27^\circ }}{{cos27^\circ }}\) | 1 |
| Inside Parentheses | \(1 - 1\) | 0 |
| Final Expression | \({0^2}\) | 0 |
| Concept | Description | Relevant Identity Example |
|---|---|---|
| Trigonometric Ratios | Ratios of sides of a right-angled triangle (sin, cos, tan, etc.) | \(\sin\theta = \frac{{Opposite}}{{Hypotenuse}}\) |
| Complementary Angles | Two angles that add up to 90° | 27° and 63° are complementary |
| Complementary Angle Identities | Relates trig ratios of complementary angles | \(\sin(90^\circ - \theta) = \cos \theta\), \(\cos(90^\circ - \theta) = \sin \theta\) |
Understanding complementary angles is very useful in trigonometry. It allows us to simplify expressions involving angles like 10° and 80°, 45° and 45°, etc. For instance, if you need to find \(\tan 80^\circ\) but only know values for angles up to 45°, you can use the identity \(\tan(90^\circ - \theta) = \cot \theta\). So, \(\tan 80^\circ = \tan(90^\circ - 10^\circ) = \cot 10^\circ\).
The co-function identities (like \(\sin \leftrightarrow \cos\), \(\tan \leftrightarrow \cot\), \(\sec \leftrightarrow \csc\)) stem directly from the concept of complementary angles in a right-angled triangle. If one acute angle is \(\theta\), the other acute angle is \(90^\circ - \theta\).
θ = 135°; γ = 15°.
What is the value of 2 cos(θ) sin(γ)?
What is the value of cot (90 + 75°)?
The value of \(\rm \frac{(cos\,9^\circ+sin \,81^\circ)(sec\,9^\circ+cosec\,81^\circ)}{cosec^2\,71^\circ+cos^2\,15^\circ-tan^2\,19^\circ+cos^2\,75^\circ} \) is:
Find the value of sin 4 30° + cos 4 30° - sin 25° cos 65° - sin 65° cos25°.
Find the value of \(cosec(60° + A) - sec(30° - A) +\frac{cosec49^{\circ}}{sec 41^{\circ}}\) .
If sin 6θ + cos 6 θ = \(\frac{1}{3}\) , 0° < θ < 90°, then what is the value of sinθ cosθ?
(secθ + tan θ )2 + \(\frac{1 + cosec θ}{1 - cosec θ'}\) 0° < θ < 90° is:
Find the value of \(\rm \frac{tan^230^\circ}{sec^230^\circ}+\frac{cosec^245^\circ}{cot^245^\circ}-\frac{sec^260^\circ}{cosec^260^\circ}\)
What is the value of sin(630 + A) + cosA?
If (A + B + C) = 90°, then what is the value of sin (A / 2) sin [(180 - B - C) / 2] + cos (A / 2) sin (B + C) / 2?
If tan 3 θ = cot(θ - 22°), where 3 θ is an angle, find the value of θ.
1 + tan 15° cot 75° is equal to:
The value of cot (-315°) is :
θ = 135°; γ = 15°.
What is the value of 2 cos(θ) sin(γ)?
If tan(90 – θ) = \(\frac{2}{\sqrt3}\), then the value of \(2\sqrt3\) tan θ + 1 is: