P and Q are acute angles such that P>Q. Then which of the following is definitely true?

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Solution

Q.
P and Q are acute angles such that P > Q. Then which of the following is definitely true

$(i) s i n P < s i n Q (i i) t a n P > t a n Q (i i i) c o s P > c o s Q (i v) c o s P > s i n Q$

1 only (i)

2 only (ii)

3 (i) and (ii) both

4 only (iv)

Ans.
(2)

Given that P and Q are acute angles such that P > Q

$L e t P = 60^{°} Q = 30 ° . C h e c k i n g (i) : W e k n o w t h a t \sin θ i n c r e a s e s a s θ i n c r e a s e s f r o m 0 ° t o 90 ° s i n 60 ° = \frac{\sqrt{3}}{2}, s i n 30 ° = \frac{1}{2} \Rightarrow s i n Q < s i n P T h u s (i) i s n o t t r u e C h e c k i n g (i i) : W e k n o w t h a t \tan θ i n c r e a s e s a s θ i n c r e a s e s f r o m 0 ° t o 90 ° \Rightarrow t a n P > t a n Q H e n c e (i i) i s t r u e . C h e c k i n g (i i i) : W e k n o w t h a t \cos θ i n c r e a s e s a s θ i n c r e a s e s f r o m 0 ° t o 90 ° S o (i i i) i s n o t t r u e . C h e c k i n g (i v) : c o s P = c o s 60 ° = \frac{1}{2}, s i n Q = s i n 30 ° = \frac{1}{2} \Rightarrow c o s P = s i n Q$

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