Video: CBSE Class X • Pack 1 • 2018 • Question 22A

CBSE Class X • Pack 1 • 2018 • Question 22A

05:12

Video Transcript

If four tan 𝜃 equals three, evaluate four sin 𝜃 minus cos 𝜃 plus one divided by four sin 𝜃 plus cos 𝜃 minus one.

Our first step here is to consider the equation four tan 𝜃 is equal to three. Dividing both sides of this equation by four gives us tan 𝜃 is equal to three-quarters.

We could approach this problem using our trigonometric identities. However, an easier way would be to use our knowledge of the Pythagorean triples. The tangent of any angle 𝜃 in a right-angled triangle is equal to the opposite divided by the adjacent.

In this case, as tan 𝜃 is equal to three-quarters or three divided by four, we know that the opposite is equal to three and the adjacent is equal to four. Using Pythagoras’s theorem, where 𝑎 squared plus 𝑏 squared is equal to 𝑐 squared, where 𝑐 is the hypotenuse, we can substitute in our values. ℎ squared is equal to three squared plus four squared. Three squared is equal to nine, and four squared is equal to 16. Adding these two values gives us ℎ squared is equal to 25. Square-rooting both sides of this equation gives us a value for ℎ, the hypotenuse, of five.

The sin of 𝜃 in any right-angled triangle is equal to the opposite divided by the hypotenuse. And the cos of angle 𝜃 is given by the adjacent divided by the hypotenuse. This means that, in our question, sin 𝜃 is equal to three-fifths and cos 𝜃 is equal to four-fifths.

Our next step is to substitute these values into the equation. This gives us four multiplied by three-fifths minus four-fifths plus one divided by four multiplied by three-fifths plus four-fifths minus one. Four multiplied by three-fifths is equal to twelve fifths. So this can be simplified to twelve fifths minus four-fifths plus one divided by twelve fifths plus four-fifths minus one. Twelve fifths minus four-fifths is equal to eight-fifths, and twelve fifths plus four-fifths is equal to sixteen fifths.

One whole one is equal to five-fifths. This can therefore be rewritten as thirteen fifths divided by eleven fifths, as eight-fifths plus five-fifths is equal to thirteen fifths. And on the bottom or denominator, sixteen fifths minus five-fifths is equal to eleven fifths. Multiplying the numerator and denominator of this fraction by five gives us an answer of thirteen elevenths, or 13 divided by 11. This means that if four tan 𝜃 is equal to three, then four sin 𝜃 minus cos 𝜃 plus one divided by four sin 𝜃 plus cos 𝜃 minus one is equal to thirteen elevenths.

As mentioned at the start, an alternative method of calculating the value of sin 𝜃 and cos 𝜃 would be to use the trigonometrical identities. Using the fact that tan squared 𝜃 plus one is equal to sec squared 𝜃 allows us to work out the value of sec squared 𝜃.

Three-quarters squared is equal to nine sixteenths. Therefore, nine sixteenths plus one is equal to sec squared 𝜃. This means that sec squared 𝜃 is equal to twenty-five sixteenths, or 25 divided by 16. As one divided by cos squared 𝜃 is equal to sec squared 𝜃, cos squared 𝜃 must be equal to sixteen twenty-fifths, or 16 over 25. Square-rooting both sides of this equation gives us a value for cos 𝜃 equal to four-fifths.

Using the identity sin squared 𝜃 plus cos squared 𝜃 is equal to one gives us the equation sin squared 𝜃 plus sixteen twenty-fifths equals one. Subtracting sixteen twenty-fifths from both sides of this equation gives us sin squared 𝜃 is equal to nine twenty-fifths. And square-rooting both sides of this equation gives us a value for sin 𝜃 equal to three-fifths. These values of sin 𝜃 and cos 𝜃 are the same as we got using the first method using Pythagoras’s theorem. This means that the final answer using this method will also be thirteen elevenths.

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