Given: cos A = 5 13 , Evaluate cot A + 1 cos A .

Consider the diagram below: cos A = 5 13 We know that, cos A = base hypotenuse ⇒ A B A C = 5 13 Therefore if length of A B = 5 x , length of A C = 13 x Since, A B 2 + B C 2 = A C 2 [Using Pythagoras Theorem] ( 5 x ) 2 + B C 2 = ( 13 x ) 2 B C 2 = 169 x 2 - 25 x 2 = 144 x 2 ∴ B C = 12 x (perpendicular) Now, cot A + 1 cos A = 5 12 + 1 5 13 = 5 12 + 13 5 = 181 60 Thus the value of cot A + 1 cos A is 181 60 .

Given: cos A = 5 13 , Evaluate cot A + 1 cos A .

Consider the diagram below: cos A = 5 13 We know that, cos A = base hypotenuse ⇒ A B A C = 5 13 Therefore if length of A B = 5 x , length of A C = 13 x Since, A B 2 + B C 2 = A C 2 [Using Pythagoras Theorem] ( 5 x ) 2 + B C 2 = ( 13 x ) 2 B C 2 = 169 x 2 - 25 x 2 = 144 x 2 ∴ B C = 12 x (perpendicular) Now, cot A + 1 cos A = 5 12 + 1 5 13 = 5 12 + 13 5 = 181 60 Thus the value of cot A + 1 cos A is 181 60 .

Given: sec A = 29 21 , Evaluate sin A - 1 tan A .

Consider the diagram below: sec A = 29 21 We know that, sec A = hypotenuse base ⇒ A C A B = 29 21 Therefore if length of A B = 21 x , length of A C = 29 x Since, A B 2 + B C 2 = A C 2 [Using Pythagoras Theorem] ( 21 x ) 2 + B C 2 = ( 29 x ) 2 B C 2 = 841 x 2 - 441 x 2 = 400 x 2 ∴ B C = 20 x (perpendicular) Now, sin A = perpendicular hypotenuse = 20 x 29 x = 20 29 tan A = perpendicular base = 20 x 21 x = 20 21 Therefore, sin A - 1 tan A = 20 29 - 1 20 21 = 20 29 - 21 20 = - 209 580 Thus the value of sin A - 1 tan A is - 209 580 .

Given: tanA = 4 3 , Evaluate cosec A cot A - sec A .

Consider the diagram below: tan A = 4 3 We know that, tan A = perpendicular base ⇒ B C A B = 4 3 Therefore if length of A B = 3 x , length of B C = 4 x Since, A B 2 + B C 2 = A C 2 [Using Pythagoras Theorem] ( 3 x ) 2 + ( 4 x ) 2 = A C 2 A C 2 = 9 x 2 + 16 x 2 = 25 x 2 ∴ A C = 5 x (hypotenuse) Now, sec A = hypotenuse base = A C A B = 5 x 3 x = 5 3 cot A = base perpendicular = A B B C = 3 x 4 x = 3 4 cosec A = hypotenuse perpendicular = A C B C = 5 x 4 x = 5 4 Therefore, cosec A cot A - sec A = 5 4 3 4 - 5 3 = - 60 44 = - 15 11 Thus the value of cosec A cot A - sec A is - 15 11 .

In a right- angled triangle, it is given that A is an acute angle and tan A = 5 12 . Find the value of cos A .

Consider the diagram below; tan A = 5 12 We know that, tan A = perpendicular base ⇒ B C A B = 5 12 Therefore if length of A B = 12 x , length of B C = 5 x Since, A B 2 + B C 2 = A C 2 [Using Pythagoras Theorem] ( 12 x ) 2 + ( 5 x ) 2 = A C 2 A C 2 = 144 x 2 + 25 x 2 = 169 x 2 ∴ A C = 13 x (hypotenuse) We know that, cos A = base hypotenuse ⇒ cos A = 12 x 13 x = 12 13 Thus the value of cos A is 12 13 .

In a right-angled triangle, it is given that A is an acute angle and tanA = 5 12 . Find the value of sin A

Consider the diagram below; tan A = 5 12 We know that, tan A = perpendicular base ⇒ B C A B = 5 12 Therefore if length of A B = 12 x , length of B C = 5 x Since, A B 2 + B C 2 = A C 2 [Using Pythagoras Theorem] ⇒ ( 12 x ) 2 + ( 5 x ) 2 = A C 2 ⇒ A C 2 = 144 x 2 + 25 x 2 = 169 x 2 ∴ A C = 13 x (hypotenuse) ⇒ sin A = perpendicular hypotenuse ⇒ sin A = 5 x 13 x = 5 13 Thus the value of sin A is 5 13 .