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1 SILVER (90+θ) & (180- θ) Sinθ & cosecθ (+ve) Rest all ( -ve ) TEA (180+θ ) & (70- θ) Tanθ & Cotθ ( +ve) Rest all ( -ve ) ALL (90- θ) & (360+θ) All positive CUPS (70+θ ) & (360-θ) Cosθ & secθ ( +ve ) Rest all ( -ve ) Sin(-θ) = - Sinθ Cos (-θ) = Cosθ Tan(-θ) = - Tanθ Cot(-θ) = - Cotθ Sec(-θ) = Secθ Cosec(-θ) = - Cosecθ In the above formulas, (-θ ) is followed the fourth quadrant of ALL SILVER TEA CUPS Sin θ + Cos θ = 1 Sec θ - Tan θ = 1 Cosec θ - Cot θ = 1 Sin (A+B) = Sin A Cos B + Cos A Sin B Sin (A-B) = Sin A Cos B - Cos A Sin B Cos (A+B) = Cos A Cos B Sin A Sin B Cos (A-B) = Cos A Cos B + Sin A Sin B Tan (A + B) = Tan (A B) = Tan A+Tan B 1 Tan A Tan B Tan A Tan B 1+Tan A Tan B X Sin X Cos X Tan X Cot X Sec X 1 Cosec X 1 Cot (A + B) = Cot (A B) = Cot A Cot B 1 Cot B+Cot A Cot A Cot B+1 Cot B Cot A Sin A = Sin A Cos A = Tan A 1+Tan A

2 Cos A = Cos A Sin A = 1- Sin A = Cos A 1 = 1 Tan A 1+Tan A Tan A = Tan A 1 Tan A Sin (A + B) Sin (A B) = Sin A Sin B = Cos B Cos A Cos (A + B) Cos (A B) = Cos A Sin B = Cos B - Sin A Tan ( π 4 Tan ( π 4 1+Tan A Cos A+Sin A + A) = = 1 Tan A Cos A Sin A 1 Tan A Cos A Sin A - A) = = 1+Tan A Cos A+Sin A Sin A = Tan A Cos A = Tan A 1+Tan A Cos A = Cos A Sin A = 1 Sin A = Cos A 1 = 1 Tan A 1+Tan A Tan A = Tan A 1 Tan A Sin 3A = 3 Sin A 4 Sin 3 A Cos 3A = 4 Cos 3 A 3 Cos A Tan 3A = 3 Tan A Tan3 A 1 3 Tan A Sin C + Sin D = Sin ( C+D C D ) Cos ( ) Sin C Sin D = Cos ( C+D C D ) Sin ( Cos C + Cos D = Cos ( C+D Cos C Cos D = - Sin ( C+D Sin 15 = Cos 75 = 1 Cos 15 = Sin 75 = + 1 Tan 15 = Cot 75 = - Cot 15 = Tan 75 = + Sin 18 = Cos 7 = 5 1 Cos 18 = Sin 7 = Tan 1 = 1 4 ) C D ) Cos ( ) C D ) Sin ( )

3 Cot 1 = + 1 Sin 36 = Cos 54 = Cos 36 = Sin 54 = Pythogoros Theorem, c = a + b (1) Can be written as b = c a () a = c b. (3) From equation (1) => c = a + b We know that, Sin θ = 1 = a + b c Therefore, 1 = Sin θ + Cos θ 1 = a + b c c 1 = ( a c ) + ( b c ) From equation, () => b = c a Opposite and Cos θ = Adjacent Hypotenuse Hypotenuse 1 = c a b 1 = c a b b 1 = ( c b ) ( a b ) We know that, Sec θ = Hypotenuse Adjacent Therefore, 1 = Sec θ Tan θ From equation (3) => a = c b 1 = c b a 1 = c b a a 1 = ( c a ) ( b a ) and Tan θ = Opposite Adjacent We know that, Cosec θ = Hypotenuse Opposite and Cot θ = Adjacent Hypotenuse

4 Therefore, 1 = Cosec θ Cot θ Fdaytalk.com Maximum Value & Minimum Values of some important Trignometric Expressions 1 asinθ + bcosθ Maximum value = a + b Minimum value = - a + b asin θ + bcos θ If a > b, Maximum value = a Minimum value = b If a < b, Maximum value = b Minimum value = a 3 atan θ + btan θ Maximum value = (Infinity) Minimum value = ab 4 asin θ + bcosec θ Maximum value = (Infinity) Minimum value = ab ax + bx + c = 0 then, Maximum value at x = b Minima and Maxima a) 1 Sin θ 1 b) 1 Cos θ 1 c) 1 Sec θ 1 d) 1 Cosec 1 e) < Tan θ < f) < Cot θ < Sin θ + Sin θ + 3 find the Maximum and Minimum values Given that, Sin θ + Sin θ + 3 Sin θ + Sin θ a (Sin θ + 1) Sin θ Sin θ (Sin θ + 1) 1 Sin θ 1

5 0 (Sin θ + 1) (Sin θ + 1) (Sin θ + 1) + 6 Therefore, Maximum value = 6 Minimum value = Fdaytalk.com Height & Distance HEIGHT & DISTANCE: AB = Height BC = Distance θ= angle x = unknown value (line of sight) Angle of Elevation:

6 Angle of Elevation, is the angle above horizontal that an observer must look see an object that is higher that the observer. Here, θ = Angle of Elevation Angle of Depression: Angle of Depression is, the angle below horizontal that an observer must look to see an object that is lower than the Observer. Here, θ = Angle of Depression From above Diagram, Sin θ = Perpendicular Hypotenus Cos θ = Base Hypotenus Tan θ = Perpendicular Base Sin θ + Cosec θ =, then find the value of sin 7 θ + Cosec 7 θ is.. Given that, Sin θ + Cosec θ = 1 Sin θ + = Sin θ We know that, x + 1 = => x = 1 x

7 Therefore, Sin θ = 1 So, Sin 7 θ + Cosec 7 θ => = (answer) Fdaytalk.com 1 Find the maximum value of 4Sinθ + 7Cosθ 4Sinθ + 7Cosθ Maximum value of asinθ + bcosθ = a + b = = = 65 = 5 (answer) Find the maximum value of 5Sin θ + Cos θ 5Sin θ + Cos θ Maximum value of asin θ + bcos θ If a > b, here 5 > Maximum value = a = 5 (answer) 3 The value of sin sin sin sin 89 0 is? Given that, sin sin sin sin 89 0 Now, sin sin sin sin to terms + sin 45 0 = (sin cos 1 0 ) + (sin cos 5 0 ) +. + to 11 terms + ( 1 ) = = 11 1 (answer) 4 Cos x + Cos y =, the value of sin x + sin y is Given that, Cos x + Cos y = Cos x 1 Cos x = 1; Cos y = 1 x = y = 0 0 [Cos 0 0 = 1] Therefore, Sin x + Sin y = 0 (answer)

8 5 If Sin 3A = Cos (A 6 0 ), where 3A is an acute angle then the value of A is Given that, Sin 3A = Cos (A 6 0 ) Cos (90 0 3A) = Cos (A 6 0 ) A = A 6 0 A = 9 0 (answer) Problems : 1 A Circus Artist is climbing a 0 m long rope, which is tightly stretched and tied from the top of a vertical pole to the ground. Find the height of the pole, if the angle made by the rope with ground level is 30 0., Given that, height of the pole AB = h meters And rope tied from the top of a vertical pole BC = 0 m From above figure, in right angle triangle CAB Sin θ = Perpendicular Hypotenus = AB BC Sin 30 0 = h 0 1 = h 0 h = 10 meters (answer) A tree breaks due to storm and the broken part bend, so that the top of the tree touches the ground making an angle 30 0 with it. The distance between the foot of the tree to the point, where the top touches the ground is 8 m. Find the height of the tree,

9 Given that from above figure, sin 30 0 = x => 1 = x.. (1) y y Cos 30 0 = 8 y => = 8 () y From equations 1 &,we get x = 8, y = 16 Therefore, length of the Tree = x + y = => 4 = 8 (answer)

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