Let, △ A B C is a right-angled triangle with ∠ B = 90 ° . Now, sin A = B C A C , cos A = A B A C and tan A = sin A cos A = B C A B . If ∠ A gets larger and larger, ∠ C gets smaller and smaller and the length of the side A B goes on decreasing. The point A gets closer to point B . When ∠ A is very close to 90 ° , ∠ C becomes very close to 0 ° and the side A C almost coincides with side B C . When ∠ C is very close to 0 ° , ∠ A is very close to 90 ° , the side A C is nearly the same as the side B C and so, sin A = B C A C is very close to 1 . Also, when ∠ A is very close to 90 ° , ∠ C is very close to 0 ° , the side A B is nearly zero, so, cos A = A B A C is very close to 0 . So, sin 90 ° = 1 , cos 90 ° = 0 Therefore, tan 90 ° = sin 90 ° cos 90 ° = 1 0 = Not defined.

Evaluate sin 30 ° + tan 45 ° - cos e c 60 ° s e c 30 ° + cos 60 ° + c o t 45 °

We have to find the value of sin 30 ° + tan 45 ° - cosec 60 ° sec 30 ° + cos 60 ° + cot 45 ° We know that sin 30 ° = 1 2 , tan 45 ° = 1 , cosec 60 ° = 2 3 sec 30 ° = 2 3 , , cos 60 ° = 1 2 , , cot 45 ° = 1 = sin 30 ° + tan 45 ° - cosec 60 ° sec 30 ° + cos 60 ° + cot 45 ° = 1 2 + 1 - 2 3 2 3 + 1 2 + 1 = 3 + 2 3 - 4 2 3 4 + 3 + 2 3 2 3 = 2 3 3 3 - 4 2 3 4 + 3 3 = 3 3 - 4 4 + 3 3 Now rationalise the denominator, = 3 3 - 4 4 + 3 3 × 4 - 3 3 4 - 3 3 = 12 3 - 27 - 16 + 12 3 16 - 27 a + b a - b = a 2 - b 2 = 24 3 - 43 - 11 = - 43 - 24 3 - 11 = 43 - 24 3 11 Hence, value of sin 30 ° + tan 45 ° - cosec 60 ° sec 30 ° + cos 60 ° + cot 45 ° is 43 - 24 3 11 .

tan A + B = 3 tan A - B = 1 3 0 ° < A + B ≤ 90 ° ; A > B find the value of A and B

Given, tan A + B = 3 ⇒ tan A + B = tan 60 ° ⇒ A + B = 60 ° - - - - - ( i ) Now, tan A - B = 1 3 ⇒ tan A - B = tan 30 ° ⇒ A - B = 30 ° - - - - - ( ii ) Adding i and ii , we get 2 A = 90 ° ⇒ A = 90 ° 2 ⇒ A = 45 ° Putting A = 45 ° in equation i , we get 45 ° + B = 60 ° ⇒ B = 60 ° - 45 ° ⇒ B = 15 ° So, A = 45 ° and B = 15 °

Evaluate 5 cos 2 60 ° + 4 sec 2 30 ° - tan 2 45 ° sin 2 30 ° + sin 2 60 °

We know that, sin 30 ° = cos 60 ° = 1 2 , cos 30 ° = 3 2 , sec 30 ° = 2 3 and tan 45 ° = 1 Now, 5 cos 2 60 ° + 4 sec 2 30 ° - tan 2 45 ° ( sin 2 30 ° + cos 2 30 ° ) = 5 × 1 2 2 + 4 2 3 2 - ( 1 ) 2 1 2 2 + 3 2 2 = 5 × 1 4 + 4 × 4 3 - 1 1 4 + 3 4 = 5 4 + 16 3 - 1 4 4 = 15 + 64 - 12 12 1 = 67 12 Hence, 5 cos 2 60 ° + 4 sec 2 30 ° - tan 2 45 ° sin 2 30 ° + cos 2 30 ° = 67 12 .