1.

O tan 48° tan 23° tan 42° tan 67° = 1(1) cos 38° cos 52° -sin 38° sin 52° = 013. Iftan 2A = cot (A-18°), where 2A is an acute angle, find the value of A.4. If tan A = cot B, prove that A + B = 90°.5. If sec 4A = cosec (A - 20°), where 4A is an acute angle, find the value of A.​

Answer»

ANSWER:

This implies that

x2+2ax=4x−4a−13

or

x2+2ax−4x+4a+13=0

or

x2+(2a−4)x+(4a+13)=0

Since the equation has just one solution instead of the usual two distinct solutions, then the two solutions MUST be same i.e. DISCRIMINANT = 0.

Hence we GET that

(2a−4)2=4⋅1⋅(4a+13)

or

4a2−16a+16=16a+52

or

4a2−32a−36=0

or

a2−8a−9=0

or

(a−9)(a+1)=0

So the values of a are −1 and 9.


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