If 2 sin θ - 1 = 0 , then prove that sec θ + tan θ = 3 .

Given, 2 sin θ - 1 = 0 We need to prove that sec θ + tan θ = 3 . 2 sin θ - 1 = 0 ⇒ sin θ = 1 2 We know that, cos θ = 1 - sin 2 θ = 1 - 1 4 = 3 2 Therefore, the value of sec θ = 1 cos θ = 2 3 And the value of tan θ = sin θ cos θ = 1 2 3 2 = 1 3 So, L H S = sec θ + tan θ = 2 3 + 1 3 = 3 3 = 3 = R H S Hence, Proved

If sin θ = 1 2 , then prove that 3 cos θ - 4 cos 3 θ = 0 .

Given, sin θ = 1 2 ⇒ θ = sin - 1 1 2 ⇒ θ = 30 ° Substituting the value of θ in the given question 3 cos θ - 4 cos 3 θ we have = 3 cos 30 ° - 4 cos 3 30 ° = 3 × 3 2 - 4 × 3 2 3 = 3 3 2 - 4 × 3 3 8 = 3 3 2 - 3 3 2 = 0 Hence, Proved.

Find an acute angle θ when, cos θ - sin θ cos θ + sin θ = 1 - 3 1 + 3 .

Given, cos θ - sin θ cos θ + sin θ = 1 - 3 1 + 3 Taking LHS and dividing numerator and denominator by cos θ , we get, 1 - sin θ cos θ 1 + sin θ cos θ = 1 - tan θ 1 + tan θ = 1 - 3 1 + 3 On comparing both the sides, we get, tan θ = 3 ⇒ θ = 60 ° Hence, the θ is 60 ° .

E M B I B E

All In One ICSE Mathematics Class 10>Chapter 18 - Trigonometric Identities>CHAPTER EXERCISE>Q 5

EASY

10th ICSE

IMPORTANT

Earn 100

With the help of the following figure, find the value of $\sec^{2} θ + \tan^{2} θ$ .

Important Questions on Trigonometric Identities

MEDIUM

10th ICSE

IMPORTANT

All In One ICSE Mathematics Class 10>Chapter 18 - Trigonometric Identities>CHAPTER EXERCISE>Q 6

If $2 \sin θ - 1 = 0$ , then prove that $\sec θ + \tan θ = \sqrt{3}$ .

MEDIUM

10th ICSE

IMPORTANT

All In One ICSE Mathematics Class 10>Chapter 18 - Trigonometric Identities>CHAPTER EXERCISE>Q 7

If $\sin θ = \frac{1}{2},$ then prove that $3 \cos θ - 4 \cos^{3} θ = 0 .$

HARD

10th ICSE

IMPORTANT

All In One ICSE Mathematics Class 10>Chapter 18 - Trigonometric Identities>CHAPTER EXERCISE>Q 8

In a right angled $∆ A C B, ∠ C = 90 °, A B = 29$ units and $B C = 21$ units. If $∠ A B C = θ,$ then find $\cos^{2} θ - \sin^{2} θ$ and $\sin^{2} θ + \cos^{2} θ .$