If sin x+cosx+tan x+cot x+sec x+cosec x=7 and sin 2x = a−b7, then ordered pair (a,b) can be

sin x+cosx+tanx+cot x+sec x+cosec x=7⇒(sin x+ cosx)+1sn x. cos x+(sin x+ cos x)sin x. cosx=7⇒(sin x + cos x)1+2sin 2x=7−2sin 2xSquaring, we have (1+ sin 2x)1+4sin22x+4sin 2x=49+4sin2 2x−28sin 2x⇒1+4sin22x+4sin 2x+sin 2x +4sin 2x+4=49+4sin22x−28sin 2x⇒sin 2x−44+36sin 2x=0⇒sin22x−44 sin 2x+36 =0 ⇒sin 2x =44±1936−1442=44±17922=22±87.sin 2x ≤1 ⇒sin 2x = 22−87. ∴a=22, b= 8.

If sin x+cosx+tan x+cot x+sec x+cosec x=7 and sin 2x = a−b7, then ordered pair (a,b) can be

sin x+cosx+tanx+cot x+sec x+cosec x=7⇒(sin x+ cosx)+1sn x. cos x+(sin x+ cos x)sin x. cosx=7⇒(sin x + cos x)1+2sin 2x=7−2sin 2xSquaring, we have (1+ sin 2x)1+4sin22x+4sin 2x=49+4sin2 2x−28sin 2x⇒1+4sin22x+4sin 2x+sin 2x +4sin 2x+4=49+4sin22x−28sin 2x⇒sin 2x−44+36sin 2x=0⇒sin22x−44 sin 2x+36 =0 ⇒sin 2x =44±1936−1442=44±17922=22±87.sin 2x ≤1 ⇒sin 2x = 22−87. ∴a=22, b= 8.

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$\begin{array}{l} I f \sin x + \cos x + \tan x + c o t x + \sec x + \cos e c x = 7 a n d \sin 2 x = a - b \sqrt{7}, t h e n o r d e r e d p a i r (a, b) c a n b e \end{array}$

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(6, 2)

(22,8)

(11, 4)

(8, 3)

answer is C.

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Detailed Solution

$\begin{array}{l} \sin x + \cos x + \tan x + \cot x + s e c x + c o s e c x = 7 \\ \Rightarrow (\sin x + \cos x) + (\frac{1}{s n x . \cos x}) + \frac{(\sin x + \cos x)}{\sin x . \cos x} = 7 \Rightarrow (\sin x + \cos x) (1 + \frac{2}{\sin 2 x}) = (7 - \frac{2}{\sin 2 x}) \\ S q u a r i n g, w e h a v e (1 + \sin 2 x) (1 + \frac{4}{\sin^{2} 2 x} + \frac{4}{\sin 2 x}) = 49 + \frac{4}{\sin^{2} 2 x} - \frac{28}{\sin 2 x} \\ \Rightarrow 1 + \frac{4}{\sin^{2} 2 x} + \frac{4}{\sin 2 x} + \sin 2 x + \frac{4}{\sin 2 x} + 4 = 49 + \frac{4}{\sin^{2} 2 x} - \frac{28}{\sin 2 x} \Rightarrow \sin 2 x - 44 + \frac{36}{\sin 2 x} = 0 \\ \Rightarrow \sin^{2} 2 x - 44 \sin 2 x + 36 = 0 \Rightarrow \sin 2 x = \frac{44 \pm \sqrt{1936 - 144}}{2} = \frac{44 \pm \sqrt{1792}}{2} = 22 \pm 8 \sqrt{7} . \\ \sin 2 x \leq 1 \Rightarrow \sin 2 x = 22 - 8 \sqrt{7} . ∴ a = 22, b = 8. \end{array}$