NDA 2017
Exam NDA
Year 2017
Subject Mathematics
Chapter Trigonometric Ratios & Identities NDA 2017 · Mathematics · Trigonometric Ratios & Identities
In PQR, R=(π)/(2) . If than ( P 2 ) and ( Q 2 ) are the roots of...
In
△ P Q R , ∠ R = π 2 \triangle \mathrm{PQR}, \angle \mathrm{R}=\frac{\pi}{2} △ PQR , ∠ R = 2 π . If than
( P 2 ) \left(\frac{\mathrm{P}}{2}\right) ( 2 P ) and
tan ( Q 2 ) a r e t h e r o o t s o f t h e e q u a t i o n \tan \left(\frac{\mathrm{Q}}{2}\right) are the roots of the equation tan ( 2 Q ) a re t h eroo t so f t h ee q u a t i o n a x²+b x+c=0
, t h e n w h i c h o n e o f t h e f o l l o w i n g i s c o r r e c t ? , then which one of the following is correct? , t h e n w hi c h o n eo f t h e f o ll o w in g i scorrec t ? Options A a = b + c \mathrm{a}=\mathrm{b}+\mathrm{c} a = b + c B b = c + a \mathrm{b}=\mathrm{c}+\mathrm{a} b = c + a C c = a + b \mathrm{c}=\mathrm{a}+\mathrm{b} c = a + b D b = c b=c b = c Correct answer C. c = a + b \mathrm{c}=\mathrm{a}+\mathrm{b} c = a + b
Step-by-step solution
∠ R = π 2 \angle \mathrm{R}=\frac{\pi}{2} ∠ R = 2 π ∴ ∠ P + ∠ Q = π 2 ⇒ ∠ P + ∠ Q 2 = π 4 … ( 1 ) tan ( P 2 + Q 2 ) = tan P 2 + tan Q 2 1 − tan P 2 tan Q 2 … ( 2 ) G i v e n , tan P 2 \therefore \angle \mathrm{P}+\angle \mathrm{Q}=\frac{\pi}{2}\Rightarrow \frac{\angle \mathrm{P}+\angle \mathrm{Q}}{2}=\frac{\pi}{4} \ldots(1)
\tan \left(\frac{\mathrm{P}}{2}+\frac{\mathrm{Q}}{2}\right)=\frac{\tan \frac{\mathrm{P}}{2}+\tan \frac{\mathrm{Q}}{2}}{1-\tan \frac{\mathrm{P}}{2} \tan \frac{\mathrm{Q}}{2}} \ldots(2)
Given, \tan \frac{\mathrm{P}}{2} ∴ ∠ P + ∠ Q = 2 π ⇒ 2 ∠ P + ∠ Q = 4 π … ( 1 ) tan ( 2 P + 2 Q ) = 1 − t a n 2 P t a n 2 Q t a n 2 P + t a n 2 Q … ( 2 ) G i v e n , tan 2 P and
tan Q 2 \tan \frac{\mathrm{Q}}{2} tan 2 Q are roots of
a x 2 + b x + c = 0 a x^2+b x+c=0 a x 2 + b x + c = 0 .
∴ tan P 2 + tan Q 2 = − b a ; tan P 2 ⋅ tan Q 2 = c a \therefore \tan \frac{\mathrm{P}}{2}+\tan \frac{\mathrm{Q}}{2}=\frac{-\mathrm{b}}{\mathrm{a}} ; \tan \frac{\mathrm{P}}{2} \cdot \tan \frac{\mathrm{Q}}{2}=\frac{\mathrm{c}}{\mathrm{a}} ∴ tan 2 P + tan 2 Q = a − b ; tan 2 P ⋅ tan 2 Q = a c .
∴ ( 2 ) ⇒ tan ( P 2 + Q 2 ) = − b a 1 − c a ⇒ tan π 4 = − b a 1 − c a \therefore(2) \Rightarrow \tan \left(\frac{\mathrm{P}}{2}+\frac{\mathrm{Q}}{2}\right)=\frac{\frac{-\mathrm{b}}{\mathrm{a}}}{1-\frac{\mathrm{c}}{\mathrm{a}}}\Rightarrow \tan \frac{\pi}{4}=\frac{\frac{-\mathrm{b}}{\mathrm{a}}}{1-\frac{\mathrm{c}}{\mathrm{a}}} ∴ ( 2 ) ⇒ tan ( 2 P + 2 Q ) = 1 − a c a − b ⇒ tan 4 π = 1 − a c a − b (from (1))
⇒ 1 = − b a − c ⇒ − b = a − c ⇒ a + b = c \Rightarrow 1=\frac{-b}{a-c} \Rightarrow-b=a-c \Rightarrow a+b=c ⇒ 1 = a − c − b ⇒ − b = a − c ⇒ a + b = c
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