Exams Joint Entrance Examination Mathematics Trigonometric Ratios and Identities Question 32 Question 32 of 298 Hard NUMERICAL
Let the set of all a ∈ R a\in R a ∈ R such that the equation cos 2 x + a sin x = 2 a − 7 \cos 2x+a\sin x=2a-7 cos 2 x + a sin x = 2 a − 7 has a solution be [ p , q ] [p,q] [ p , q ] and r = tan 9 ∘ − tan 27 ∘ − 1 cot 63 ∘ + tan 81 ∘ r=\tan {9}^{\circ }-\tan {27}^{\circ }-\frac{1}{\cot {63}^{\circ }}+\tan {81}^{\circ } r = tan 9 ∘ − tan 27 ∘ − c o t 63 ∘ 1 + tan 81 ∘ , then pqr is equal to \_\_\_\_\_\_\_\_\_\_\_\_.
Answer: 48
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cos 2 x + a ⋅ sin x = 2 a − 7 a ( sin x − 2 ) = 2 ( sin x − 2 ) ( sin x + 2 ) sin x = 2 , a = 2 ( sin x + 2 ) ⇒ a ∈ [ 2 , 6 ] p = 2 q = 6 r = tan 9 ∘ + cot 9 ∘ − tan 27 − cot 27 r = 1 sin 9 ⋅ cos 9 − 1 sin 27 ⋅ cos 27 = 2 [ 4 5 − 1 − 4 5 + 1 ] r = 4 p ⋅ q ⋅ r = 2 × 6 × 4 = 48 \begin{gathered} \cos 2x+a\cdot \sin x=2a-7 \\ a(\sin x-2)=2(\sin x-2)(\sin x+2) \\ \sin x=2,a=2(\sin x+2) \\ \Rightarrow a\in [2,6] \\ p=2\, q=6 \\ r=\tan {9}^{\circ }+\cot {9}^{\circ }-\tan 27-\cot 27 \\ r=\frac{1}{\sin 9\cdot \cos 9}-\frac{1}{\sin 27\cdot \cos 27} \\ =2[\frac{4}{\sqrt{5}-1}-\frac{4}{\sqrt{5}+1}] \\ r=4 \\ p\cdot q\cdot r=2\times 6\times 4=48 \end{gathered} cos 2 x + a ⋅ sin x = 2 a − 7 a ( sin x − 2 ) = 2 ( sin x − 2 ) ( sin x + 2 ) sin x = 2 , a = 2 ( sin x + 2 ) ⇒ a ∈ [ 2 , 6 ] p = 2 q = 6 r = tan 9 ∘ + cot 9 ∘ − tan 27 − cot 27 r = sin 9 ⋅ cos 9 1 − sin 27 ⋅ cos 27 1 = 2 [ 5 − 1 4 − 5 + 1 4 ] r = 4 p ⋅ q ⋅ r = 2 × 6 × 4 = 48 Answer the question to unlock