Introduction to Trigonometry | Chapter Assessment | English | Grade 10

If cot θ = 40/9, then the value of cosec θ is ____

• cosec θ = 9 / 41

• cosec θ = 41/ 9

• cosec θ = 40/ 9

• cosec θ = 9/ 40

Answer explanation

Consider a right angled triangle. If cot θ = 40 / 9. Then, let side opposite to ∠θ = 9x, Adjacent side = 40x Calculating Hypotenuse using Pythagoras Theorem, Hypotenuse²= ( 9x)² + (40x)² Hypotenuse²= 81x² + 1600x² = 1681x² Hypotenuse = √1681x² = 41x cosec θ = Hypotenuse/ Side opposite to ∠θ So cosecθ = 41x / 9x = 41/ 9 So the correct answer is Option 2

Tags

CCSS.HSF.TF.C.8

If 4 tan A = 3, then find the value of of the following expression:

• 1/21

• 23/41

• ─1/21

• Cannot be determined

Answer explanation

If A = 30⁰, verify that 3cosA ─ 4cos³A = 0 Click on 'Yes' after verifying.

• Yes

• No

Answer explanation

Given: A = 30⁰ 3 cos A ─ 4 cos³ A =0 Consider LHS = 3 cos A ─ 4 cos³ A = 3 (√3/2) - 4(√3/2)³ = (3√3)/2 ─ 4.(3√3/8) = 3√3/2 ─ 3√3/2 = 0 Therefore, LHS = RHS Hence, verified.

Evaluate the given expression: cot² 30⁰ ─ 2cos² 60⁰ ─ 3/4sec² 45⁰ ─ 4sec² 30⁰

• ─7

• ─ 31

• ─ 13 / 3

Answer explanation

The values of -- cot 30⁰ = √3 cos 60⁰ = 1/2 sec 45⁰ = √2 sec 30⁰ =2 /√ 3 Substituting these values in the given expression, we get, cot² 30⁰ ─ 2cos² 60⁰ ─ 3/4 sec² 45⁰ ─ 4sec² 30⁰ = (√3)² ─ 2 (1/4) ─ 3/4 (√2)² ─ 4 (2/√3)² = 3 ─ 1/2 ─3/2 ─16/3 = 3 ─ 2 ─16/3 = 1 ─16/3 = ─13/3 So the correct answer is Option 3.

If cosec (A+B) = 1 and sec (A − B) = 2, 0⁰ < A + B ≤ 90⁰, A > B, then find the values of A and B.

• A = 75⁰ B = 15⁰

• A = 45⁰ B = 45⁰

• A = 15⁰ B = 75⁰

• Cannot be determined

Answer explanation

cosec (A+B) = 1 (Given) So, A + B = 90⁰ … (1) ( as cosec 90⁰ = 1) sec (A ─ B) = 2 (Given) So, A ─ B = 60⁰ ….. (2) ( as sec 60⁰ = 2) Adding Eq.1 and Eq. 2 together, 2A = 150⁰ A = 75⁰ Substituting the value of A in Eq.1, 75⁰ + B = 90⁰ B = 15⁰ So, the correct answer is Option 1

Tags

CCSS.HSG.SRT.C.7

State whether the following equation is true or false. sin⁸θ ─ cos⁸θ = (sin²θ ─ cos²θ)( 1 ─ 2sin²θcos²θ)

• TRUE

• FALSE

Answer explanation

To Prove: sin⁸θ ─ cos⁸θ = (sin²θ ─ cos²θ)( 1 ─2sin²θcos²θ) Proof: Consider LHS = sin⁸θ ─ cos⁸θ = (sin⁴θ + cos⁴θ)(sin⁴θ ─ cos⁴θ) ...........{Since a²─ b² = (a+b)(a─b)} = (sin⁴θ + cos⁴θ)(sin²θ ─ cos²θ)(sin²θ + cos²θ) = (sin⁴θ + cos⁴θ)(sin²θ ─ cos²θ) …......{Since sin²θ + cos²θ= 1} Now consider, (sin²θ + cos²θ)² = sin⁴θ + cos⁴θ + 2sin²θcos²θ So, sin⁴θ + cos⁴θ = (sin²θ + cos²θ)² ─ 2sin²θcos²θ = 1 ─ 2sin²θcos²θ Hence, (sin⁴θ + cos⁴θ)(sin²θ ─ cos²θ) can be written as, (sin⁴θ + cos⁴θ)(sin²θ ─ cos²θ) = (sin²θ ─ cos²θ) (1 ─ 2sin²θcos²θ) This is equal to RHS Hence the equation is true and the correct answer is Option 1

If 7cosecA ─ 3cotA = 7, then prove that 7cotA ─ 3cosecA = 3 Click on 'Yes' after completing the proof.

• Yes

• No

Answer explanation

7cosecA ─ 3cotA = 7, Rewrite it as 7cosecA ─ 7 = 3cotA 7(cosecA - 1) = 3cotA Multiplying both sides with (cosecA + 1). 7(cosecA - 1)(cosecA + 1) = 3cotA(cosecA + 1) 7(cosec²A - 1) = 3cotA(cosecA + 1) 7 cot²A= 3cotA(cosecA + 1) .......( As cosec²A -1 = cot²A) On dividing both the sides with cotA, we get, 7cotA = 3(cosecA + 1) 7cotA = 3cosecA + 3 7cotA ─3cosecA = 3 Hence Proved