In a △ A B C if cot A 2 , cot B 2 , cot C 2 are in A . P , then prove that a , b , c are in A . P .

Given, cot A 2 , cot B 2 , cot C 2 are in AP . ⇒ 2 cot B 2 = cot A 2 + cot C 2 We know that, half angle formula of cot A 2 is obtained from sin A 2 cos A 2 . ⇒ 2 s ( s - b ) Δ = s ( s - a ) Δ + s ( s - c ) Δ ⇒ 2 ( s - b ) = ( s - a ) + ( s - c ) ⇒ a - b + c = 2 s - a - c ⇒ a + c - b = a + b + c - a - c ⇒ a + c = 2 b ∴ a , b , c are in AP Hence, proved.

In a △ A B C if sin 2 A 2 , sin 2 B 2 , sin 2 C 2 are in H . P , then show that a , b , c are in H . P .

Given, sin 2 A 2 , sin 2 B 2 , sin 2 C 2 are in HP . i.e., ( s - b ) ( s - c ) b c , ( s - c ) ( s - a ) a c , ( s - a ) ( s - b ) a b are in HP ⇒ b c ( s - b ) ( s - c ) , a c ( s - a ) ( s - c ) , a b ( s - a ) ( s - b ) are in AP Multiplying with ( s - a ) ( s - b ) ( s - c ) a b c , we have b c ( s - a ) ( s - b ) ( s - c ) a b c ( s - b ) ( s - c ) , a c ( s - a ) ( s - b ) ( s - c ) a b c ( s - a ) ( s - c ) , a b ( s - a ) ( s - b ) s - c a b c ( s - a ) ( s - b ) ⇒ s - a a , s - b b , s - c c are in AP Adding 1 to every term, we have s - a a + 1 , s - b b + 1 , s - c c + 1 are in AP s a , s b , s c are in AP ⇒ 1 a , 1 b , 1 c are in AP ⇒ a , b , c are in HP Hence, proved.

In a △ A B C if C = 90 ° , then prove that a 2 + b 2 a 2 - b 2 sin A - B = 1 .

Given that, ∆ A B C , ∠ C = 90 ° Then, A + B + C = 180 ° ⇒ A + B = 90 ° ...(i) From sine rule, a sin A = b sin B = c sin C = k To prove, a 2 + b 2 a 2 - b 2 sin A - B = 1 LHS = a 2 + b 2 a 2 - b 2 sin A - B = 1 = ( k sin A ) 2 + ( k sin B ) 2 ( k sin A ) 2 - ( k sin B ) 2 sin A - B = sin 2 A + sin 2 B sin 2 A - sin 2 B sin A - B [Cancelling k from numerator and denominator] = 1 - cos 2 A 2 + 1 - cos 2 B 2 1 - cos 2 A 2 - 1 - cos 2 B 2 sin A - B = 1 - cos 2 A + 1 - cos 2 B 1 - cos 2 A - 1 + cos 2 B sin A - B = 2 - cos 2 A - cos 2 B cos 2 B - cos 2 A sin A - B = 2 - ( cos 2 A + cos 2 B ) cos 2 B - cos 2 A sin A - B = 2 - 2 cos ( 2 A + 2 B ) 2 cos ( 2 A - 2 B ) 2 2 sin ( 2 B + 2 A ) 2 sin ( 2 A - 2 B ) 2 sin A - B = 2 - ( 2 cos ( A + B ) cos ( A - B ) ) 2 sin ( A + B ) sin ( A - B ) sin A - B = 2 - ( 2 cos ( A + B ) cos ( A - B ) ) 2 sin ( A + B ) sin ( A - B ) sin A - B = 2 - ( 2 cos ( A + B ) cos ( A - B ) ) 2 sin ( A + B ) From (i), we get = 2 - 2 cos 90 ° cos ( A - B ) 2 sin 90 ° = 2 - 0 2 = 1 = RHS Hence, proved.