 ×     #### Thank you for registering.

One of our academic counsellors will contact you within 1 working day.

Click to Chat

1800-1023-196

+91-120-4616500

CART 0

• 0

MY CART (5)

Use Coupon: CART20 and get 20% off on all online Study Material

ITEM
DETAILS
MRP
DISCOUNT
FINAL PRICE
Total Price: Rs.

There are no items in this cart.
Continue Shopping
```
if alpha and beta are the roots of the equation 3cosx+4sinx=5 then cot(alpha+beta) equals

```
9 months ago

```							3cosx + 4sinx = Asin(x + Ф) Multiply and divide with = 5 in LHSe.g., (3cosx + 4sinx)/5 = Asin(x + Ф)⇒5(3/5.cosx + 4/5.sinx) = Asin(x + Ф) Let cosФ= 3/5 then, sinФ = 4/5Now, tanФ = sinФ/cosФ = 4/3⇒ Ф = tan⁻¹(4/3) ----(1) And 5(3/5.cosx + 4/5.sinx)= 5(sinФ.cosx + cosФ.sinx) = Asin(x + Ф)⇒ 5sin(x + Ф) = Asin(x + Ф)From equation (1),⇒ 5sin(x + tan⁻¹(4/3)) = Asin(x + Ф)Compare both sides,A = 5 and Ф = tan⁻¹(4/3)
```
9 months ago Saurabh Koranglekar
10233 Points
```							Dear studentThe simplification isSin ( x + 37 ) = 1= sin ( 90+ 2n*pi)You can solve accordingly by evaluating alpha and betaRegards
```
9 months ago
```							note that aruns soln doesnt answer the ques.we use the standard formulas:cosx= (1 – t^2)/(1+t^2) and sinx= 2t/(1+t^2) where t= tan(x/2).so eqn becomes3(1 – t^2)+4(2t)= 5(1+t^2)or 8t^2 – 8t + 2= 0 or 4t^2 – 4t + 1= 0or (2t – 1)^2= 0or t= tan(x/2)= ½ so that tan(p)= tan(q)= ½ where p= alpha/2 and q= beta/2.cot(alpha+beta)= 1/tan(alpha+beta)= 1/tan(2p + 2q)= (1 – tan2p*tan2q)/(tan2p + tan2q)now, tan2p= 2tanp/(1 – tan^2p)= 4/3. similarly tan2q= 4/3hence, cot(alpha+beta)= (1 – tan2p*tan2q)/(tan2p + tan2q)= – 7/24kindly approve :))
```
9 months ago
Think You Can Provide A Better Answer ?

## Other Related Questions on Trigonometry

View all Questions »  ### Course Features

• 731 Video Lectures
• Revision Notes
• Previous Year Papers
• Mind Map
• Study Planner
• NCERT Solutions
• Discussion Forum
• Test paper with Video Solution  ### Course Features

• 31 Video Lectures
• Revision Notes
• Test paper with Video Solution
• Mind Map
• Study Planner
• NCERT Solutions
• Discussion Forum
• Previous Year Exam Questions