Question

If sinx + COSX + tanx + cot x + secx + cosecx = 7 and sin2x =a -b√7, then
ordered pair (a, b) can be,
(A) (6,2)
(B) (8,3)
(C) (22, 8)
(D) (11, 4)​

Answer 1

option second may be the correct answer

Answer 2

Answer:

Step-by-step explanation:

sinx+cosx+tanx+secx+cscx+cotx=7  

⟹sinx+cosx+sinx/cosx+1/cosx+1/sinx+cosx/sinx=7  

⟹(sinx+1cosx)+(cosx+1sinx)+(sinxcosx+cosxsinx)=7  

⟹sinxcosx+1cosx+sinxcosx+1sinx+1sinxcosx=7—(A)  

Let  sin2x=2sinxcosx=p⟹sinxcosx=p2  

(A)⟹1+p2cosx+1+p2sinx+2p=7  

⟹(1+p2)(sinx+cosxsinxcosx+2p=7  

⟹(2+p)((sinx+cosx)p=7p−2p  

Assuming  p=sin2x2≠0:  

(sinx+cosx)=7p−2p+2  

Squaring both sides:

sin2x+cos2x+2sinxcosx=(7p−2p+2)2=49p2−28p+4p2+4p+4  

⟹1+p=p2+4p+4+48p2−32pp2+4p+4=1+48p2−32pp2+4p+4  

⟹p=48p2−32pp2+4p+4  

Since  p≠0,1=48p−32p2+4p+4  

⟹p2+4p+4−48p+32=0  

⟹p2−44p+36=0  

⟹p=44±442−(4∗1∗36−−−−−−−−−−−−−√2=44±167–√2  

⟹p=sin(2x)=22±8√7

∴a=22 and b=8

(C)