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`        If 5A and 9A are supplementary angles then find the exact value of (4 cos2A - sec4A)`
one month ago

Saurabh Koranglekar
8380 Points
```							Dear student5A +9A = 180A= 180/14Or pi/14Exact value of given equation is 2Regards
```
one month ago
Vikas TU
11761 Points
```							Here,given 5A and 9A are supplementary angles;So, 5A+9A=π14A=πThen, divided by 2 we get,7A =π/2,it is also written as;4A +3A =π/24A =π/2 –3ABy taking cosine on both side ,we getcos(4A)=cos(π/2 −3A)This equation is satisfied by A=π/14,3π/14,5π/14 ….2cos²(2A)−1=sin(3A)By using formula 1–cos2A =2sin²A=>2[1–2sin²A]²−1=[3sinA–4sin³3A]Let x=sinA=>2(4x⁴−4x²+1)−1=3x–4x³=>8x⁴−8x²+2–1=3x–4x³=>8x⁴+4x³−8x²–3x +1=0=>8x⁴–8x²+4x³–3x+1=0=>8x²(x²–1)+4x(x²–1) +x+1=0=>(x²–1)(8x²+4x)+(x+1) =0=>(x+1)(x–1)(8x²+4x)+(x+1)=0(x+1){(x–1)(8x²+4x) +1}=0As you can see at x=−1,the above equation are satisfy and (x+1) is a one of its factor other 3 values of x possible.So,we consider here only x =–1=sinAThen,by putting x=sinA, we get the value of4cos2A–sec4A =4(1–2sin²A) - 1/(cos4A)Because secA = 1/cosA=4(1–2sin²A) - 1/(2cos²2A –1)=4(1–2sin²A) - 1/{2(1–2sin²A)²–1}=4{1–2(-1)²} - 1/{2(1–2×-1²)²–1}=4(1–2) - 1/{2(1–2)²–1}=–4 - 1/(2–1)=–4 –1–5,Hence, the value 4cos2A –sec4A = –5,Ans.You try to find other values of 4cos2A–sec4AI hope it helps you…!!!
```
one month ago
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### Course Features

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