2024年12月13日 · Ex 8.1, 8 If 3 cot A = 4, check whether ( (1 − 𝑡𝑎𝑛2𝐴))/ ( (1 + 𝑡𝑎𝑛2𝐴))= cos2 A – sin2A or not. Given 3 cot A = 4 cot A = 𝟒/𝟑 So, tan A = 1/cot⁡𝐴 tan A = 1/ ( (4/3) ) tan A = 𝟑/𝟒 Now, tan A = 3/4 (𝑺𝒊𝒅𝒆 𝒐𝒑𝒑𝒐𝒔𝒊𝒕𝒆 ∠𝑨)/ (𝑺𝒊𝒅𝒆 ...

3 cot A = 4. Thus, cot A = 4/3. Let ΔABC be a right-angled triangle where angle B is a right angle. cot A = side adjacent to ∠A / side opposite to ∠A = AB/BC = 4/3. Let AB = 4k and BC = 3k, where k is a positive integer. By applying the Pythagoras theorem in ΔABC, we get, AC 2 = AB 2 + BC 2. = (4k) 2 + (3k) 2. = 16k 2 + 9k 2. = 25k 2. AC = √ 25k².

If 3 cot A = 4, where 0° < A < 90°, then sec A is equal to 5 4. Explanation: 3 cot A = 4 cot A = 4 3 ⇒ tan A = 1 cot A = 3 4 As we know, sec A = 1 + (tan A) 2 = 1 + (3 4) 2 = 1 + 9 16 = 16 + 9 16 …

2023年12月25日 · cosec (180° – θ) = – cosec θ Quadrants of trigonometry If 3 cot A = 4, check whether (1 – tan2A)/ (1 + tan2A) = cos2A – sin2A or not. Solution: If 3 cot A = 4 therefore cot A = 4/3 tan A = 3/4 to prove (1 – tan2A)/ (1 + tan2A) = cos2 A – sin2 A Take LHS (1 – tan2A)/ (1 + tan2A) = [ {1 – (3/4)2 }] / { [ 1 + (3/4)2}]

2024年9月10日 · If \ (3 \cot A=4,\) where \ (0^ {\circ}<A<90^ {\circ},\) then sec A is equal to (A) \ (\frac {5} {4}\) (B) ... ) (C) \ (\frac {5} {3}\) (D) \ (\frac {3} {4}\)

Exercise 8.1 class 10 If 3 cot A =4 check whether (1-tan²A/ (1+tan²A) = cos²A–sin²A Hint💡: Transpose 3 to RHS cot A = 3 by 4 or 3/4 tan²A = tan A x tan A Remember: sin A = P/H cos A...

2024年11月8日 · To solve the first part, we start with the equation given: 3 cot A = 4. From this, we can find cot A = 4/3. Using the identity, we know that cosec^2 A = 1 + cot^2 A. Therefore, cosec^2 A = 1 + (4/3)^2 = 1 + 16/9 = 25/9. Now we can substitute this value into the expression we need to evaluate: (cosec^2 A + 1) / (cosec^2 A - 1).

2024年11月22日 · To find secA given 3cotA=4, we start by solving for cotA. Then, we use the Pythagorean identity to find secA. Given: 3cotA=4. Solve for cotA: cotA= 34. Use the identity …

It is given that 3cot A = 4 or cot A = 4 3. Consider a right triangle ABC, right-angled at point B. cot A = Side adjacent to ∠A Side opposite to ∠A Side adjacent to ∠A Side opposite to ∠A. AB BC …

Given, 3 cot A = 4 ⇒ cot A = 4 3 4 3 By definition, tan A = 1 C ot A 1 C o t A = 1 (4 3) 1 (4 3) ⇒ tan A = 3 4 3 4 Thus, Base side adjacent to ∠A = 4 Perpendicular side opposite to ∠A = 3 In ΔABC, Hypotenuse is unknown Thus, by applying Pythagoras theorem in ΔABC We get AC2 = AB2 + BC2 AC2 = 42 + 32 AC2 = 16 + 9 AC2 = 25 AC = √25 AC = 5