2024年12月13日 · 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 𝒂𝒅𝒋𝒂𝒄𝒆𝒏𝒕 ∠𝑨) = 𝟑/𝟒 𝑩𝑪/𝑨𝑩 = 𝟑/𝟒 Let BC = 3x & AB = 4x We find AC using Pythagoras theorem In right ...

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.

In ΔPQR, right angled at Q, PQ = 4 cm and RQ = 3 cm. Find the values of sin P, sin R, sec P and sec R. If ∠A and ∠P are acute angles such that tan A = tan P, then show that ∠A = ∠P. If θ = 30° verify that `sin 2theta = (2 tan theta)/(1 + tan^2 theta)`

If 3 cot A = 4, where 0° < A < 90°, then sec A is equal to `bbunderline(5/4)`. Explanation: 3 cot A = 4. cot A = `4/3` ⇒ tan A = `1/cot A = 3/4` As we know, sec A = `sqrt(1 + ( tan A )^2` = `sqrt(1 + ( 3/4 )^2` = `sqrt (1 + 9/16)` = `sqrt(16 + 9)/16` = `sqrt25/16` = `5/4`

2024年9月10日 · Correct option is (A) \(\frac{5}{4}\) \(3 \cot A = 4\) \( \cot A = \frac 43\) \(\Rightarrow \tan A = \frac 1 {\cot A} = \frac 34\) As we know that, \(\sec A = \sqrt {1 + (\tan A)^2}\) \(= \sqrt { 1 + (\frac 34)^2}\) \(= \sqrt { 1 + \frac 9 {16}}\) \(= \sqrt { \frac {16+9} {16}}\) \(= \sqrt { \frac {25} {16}}\) \(= { \frac {5} {4}}\)

2023年12月25日 · If 3 cot A = 4, check whether (1 – tan 2 A)/(1 + tan 2 A) = cos 2 A – sin 2 A or not. Solution: If 3 cot A = 4. therefore cot A = 4/3. tan A = 3/4. to prove (1 – tan 2 A)/(1 + tan 2 A) = cos 2 A – sin 2 A Take LHS (1 – tan 2 A)/(1 + tan 2 A) = [{1 – (3/4) 2}] / { [ 1 + (3/4) 2}] = [{1 – 9/16}] / {[ 1 + 9/16}] = {(16 -9)/16 ...

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).

3 cot A = 4 . ⇒ cot A = \(\frac{4}{3}\) By definition, tan A = \(\frac{1}{Cot\, A}\) = \(\frac{1}{({4\over3})}\) ⇒ tan A = \(\frac{3}{4}\) 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 . AC 2 = AB 2 + BC 2 . AC 2 = 4 2 ...

2023年11月28日 · 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, AC2 = AB2 + BC2 = (4k)2 + (3k)2 = 16k2 + 9k2 = 25k2. AC = √25k² = 5k. Therefore, tan A = side opposite to ∠A / side adjacent to ∠A = BC/AB = 3k/4k = 3/4