Let a line OP start from OX revolve round O in the anticlockwise direction. Then it will trace positive angle XOP. If OP is in the first quadrant, then \(\angle\)XOP lies between 0° and 90°. If OP is in the second quadrant, then \(\angle\)XOP lies between 90° and 180°. Similarly, when OP is in the third or the fourth quadrant, then \(\angle\)XOP lies between 180° and 270° or 270° and 360° respectively.If OP revolves in the clockwise direction, the angle traced is negative.
In the above figure, lines measured along OX' or OY' are considered positive. Lines measured along OX and OY are considered negative. If PM is perpendicular to XX', then
Thus the values of trigonometric ratios in different quadrants are as follows:-
1st quadrant | 2nd quadrant | 3rd quadrant | 4th quadrant |
all the ratios are (+e) positive | sin and cosec (+ve) remaining (-ve) | tan and cot (+ve) remaining (-ve) | cos and sec (+ve) remaining (-v) |
Ratios of (90° - A)
Let revolving line OP start from OX and trace out an angle XOP = A. Let another line OQ (= OP) revolve from OX to OYn and return back to OQ through an angle YOQ = A.
Then \(\angle\)XOQ = 90° - A
Draw perpendiculars PM and QN from P and Q respectively to OX.
Now, in \(\triangle\)^{s} ONQ and OMP, \(\angle\)OQN = \(\angle\)MOP, \(\angle\)ONQ \(\angle\)OMP and OQ = OP. So, triangles ONQ and OMP are congruent.
Corresponding sides of a congruent triangle are equal.
\(\therefore\) OQ = OP, NQ = OM and ON = MP.
Now,
sin(90° - A) = \(\frac{NQ}{OQ}\)= \(\frac{OM}{OP}\) = cos A
cos(90° - A)= \(\frac{ON}{OQ}\)= \(\frac{MP}{OP}\) = sin A
tan(90° - A) = \(\frac{NQ}{ON}\)= \(\frac{OM}{MP}\) = cot A
cosec(90° - A)= \(\frac{OQ}{NQ}\)= \(\frac{OP}{OM}\) = sec A
sec(90° - A)= \(\frac{OQ}{ON}\)= \(\frac{OP}{MP}\) = cosec A
cot(90° - A)= \(\frac{ON}{NQ}\)= \(\frac{MP}{OM}\) = tan A
Ratios of (90° + A)
Let a revolving line OP start from OX and trace out an angle XOP = A. Let another line OQ (=OP) revolve from OX to OY and agian to OQ through an angle YOQ = A.
Then, \(\angle\)XOQ = 90° + A
Draw perpendicular PM and QN from P and Q respectively to XOX'. In \(\triangle\)^{s} ONQ and OMP, OQ = OP, \(\angle\)OQN = \(\angle\)MOP and \(\angle\)ONQ =\(\angle\)OMP. So, \(\triangle\) ONQ ≅\(\triangle\)OMP.
Corresponding sides of congruent triangles are equal.
So, OQ = OP, ON = MP and NQ = OM.
By the rule of signs, OQ = OP, ON = -MP and NQ = OM.
Now,
sin(90° + A) = \(\frac{NQ}{OQ}\)= \(\frac{OM}{OP}\) = cos A
cos(90° + A)= \(\frac{ON}{OQ}\)= \(\frac{-MP}{OP}\) = -sin A
tan(90° + A) = \(\frac{NQ}{ON}\)= \(\frac{OM}{-MP}\) = -cot A
cosec(90° + A)= \(\frac{OQ}{NQ}\)= \(\frac{OP}{OM}\) = sec A
sec(90° + A)= \(\frac{OQ}{ON}\)= \(\frac{OP}{-MP}\) = -cosec A
cot(90° + A)= \(\frac{ON}{NQ}\)= \(\frac{-MP}{OM}\) = -tan A
Ratios of (180° - A)
Let a revolving line OP start from OX and trace out an angle XOP = A. Let another line OP (= OP) revolve from OX to OX' and return to OQ through an angle X'OQ = A.
Then, \(\angle\)XOQ = 180° - A.
Draw perpendicular PM and QN from P and Q respectively to XOX'.
In \(\triangle\)^{s}ONQ and OMP, \(\angle\)ONQ =\(\angle\)OMP,\(\angle\)QON =\(\angle\)POM and OQ = OP. So, \(\triangle\)ONQ≅ \(\triangle\)OMP.
Corresponding sides of congruent triangles are equal.
\(\therefore\)OQ = OP, ON = OM and NQ = MP.
By the rule of signs, OQ = OP, ON = -OM and NQ = MP
Now,
sin(180° - A) = \(\frac{NQ}{OQ}\)= \(\frac{MP}{OP}\) =sin A
cos(180° - A)= \(\frac{ON}{OQ}\)= \(\frac{-OM}{OP}\) = -cos A
tan(180° - A) = \(\frac{NQ}{ON}\)= \(\frac{MP}{-OM}\) = -tan A
cosec(180° - A)= \(\frac{OQ}{NQ}\)= \(\frac{OP}{MP}\) = cosec A
sec(180° - A)= \(\frac{OQ}{ON}\)= \(\frac{OM}{-OM}\) = -sec A
cot(180° - A)= \(\frac{ON}{NQ}\)= \(\frac{-OM}{MP}\) = -cot A
Ratios of (180° + A)
Let a revolving line OP start from OX and trace out an angle XOP = A. Let another line OQ (= OP) revolve from OX to OX' and again to OQ through an angle X'OQ = A.
Then, \(\angle\)XOQ = 180° + A.
Draw perpendiculars PM and QN from P and Q respectively to XOX'.
In \(\triangle\)^{s}ONQ and OMP, \(\angle\)ONQ =\(\angle\)OMP,\(\angle\)QON =\(\angle\)POM and OQ = OP.
\(\therefore\) \(\triangle\)ONQ = \(\triangle\)OMP.
Corresponding sides of congruent triangles are equal.
\(\therefore\) OQ = OP, ON = OM and NQ = MP.
By the rule of signs, OQ = OP, ON = -OM and NQ = -MP.
Now,
sin(180° + A) = \(\frac{NQ}{OQ}\)= \(\frac{-OM}{OP}\) = -sin A
cos(180° + A)= \(\frac{ON}{OQ}\)= \(\frac{-OM}{OP}\) = -cos A
tan(180° + A) = \(\frac{NQ}{ON}\)= \(\frac{-MP}{-OM}\) = tan A
cosec(180° + A)= \(\frac{OQ}{NQ}\)= \(\frac{OP}{-MP}\) = -cosec A
sec(180° +A)= \(\frac{OQ}{ON}\)= \(\frac{OP}{-OM}\) = -sec A
cot(180° +A)= \(\frac{ON}{NQ}\)= \(\frac{-OM}{-MP}\) = cot A
Ratios of (270° - A)
Let a revolving line OP start from OX and trace out an angle XOp = A. Let another line OQ = (= OP) revolve from OX to OY' and return to OQ through an angle Y'OQ = A.
Then, \(\angle\)XOQ = 270° - A.
Draw perpendicular PM and QN from P and Q respectively to XOX'.
In \(\triangle\)^{s}ONQ and OMP, \(\angle\)ONQ =\(\angle\)OMP,\(\angle\)OQN =\(\angle\)MOP ans OQ = OP.
\(\therefore\) \(\triangle\)ONQ ≅\(\triangle\)OMP.
Corresponding sides of congruent triangles are equal.
So, OQ = OP, ON = MP, NQ = OM
By the rule of signs, OQ= OP, ON = -MP and NQ = -OM.
Now,
sin(270° - A) = \(\frac{NQ}{OQ}\)= \(\frac{-OM}{OP}\) = -cos A
cos(270° -A)= \(\frac{ON}{OQ}\)= \(\frac{-MP}{OP}\) = -sin A
tan(270° - A) = \(\frac{NQ}{ON}\)= \(\frac{-OM}{-MP}\) = cot A
cosec(270° - A)= \(\frac{OQ}{NQ}\)= \(\frac{OP}{-OM}\) = -sec A
sec(270° -A)= \(\frac{OQ}{ON}\)= \(\frac{OP}{-MP}\) = -cosec A
cot(270° - A)= \(\frac{ON}{NQ}\)= \(\frac{-MP}{-OM}\) = tan A
Ratios of (270° + A)
Let s revolving
line OP start from OX and trace out an angle XOP = A. Let another line OQ = (- OP) revolve from OX to OY' and return to OQ through an angle Y'OQ = A.
Then, \(\angle\)XOQ = 270° - A.
Draw perpendicular PM and QN from P and Q respectively to XOX'.
In \(\triangle\)^{s}ONQ and OMP, \(\angle\)ONQ =\(\angle\)OMP,\(\angle\)OQN =\(\angle\)MOP ans OQ = OP.
\(\therefore\) \(\triangle\)ONQ ≅\(\triangle\)OMP.
Corresponding sides of congruent triangles are equal.
So, OQ = OP, ON = MP, NQ = OM
By the rule of signs, OQ= OP, ON = -MP and NQ = -OM.
Now,
sin(270° +A) = \(\frac{NQ}{OQ}\)= \(\frac{-OM}{OP}\) = -cos A
cos(270° +A)= \(\frac{ON}{OQ}\)= \(\frac{MP}{OP}\) = sin A
tan(270° +A) = \(\frac{NQ}{ON}\)= \(\frac{-OM}{MP}\) = -cot A
cosec(270° +A)= \(\frac{OQ}{NQ}\)= \(\frac{OP}{-OM}\) = -sec A
sec(270° + A)= \(\frac{OQ}{ON}\)= \(\frac{OP}{MP}\) = cosec A
cot(270° +A)= \(\frac{ON}{NQ}\)= \(\frac{MP}{-OM}\) = -tan A
Ratios of (360° - A)
Let a revolving line OP start from OX and trace out an angle XOP = A. Let another line OQ = OP revolve from OX to OX and return back to OQ through an angle XOQ = A.
Then, \(\angle\)XOQ = 360° - A.
Draw perpendiculars PM and QN from P and Q respectively to OX.
In \(\triangle\)sONQ and OMP, \(\angle\)ONQ = \(\angle\)OMP, \(\angle\)NOQ = \(\angle\)MOP and OQ = OP.
\(\therefore\) \(\triangle\) ONQ≅ \(\triangle\) OMP.
Corresponding sides of congruent triangles are equal.
So, OQ = OP, ON = OM and NQ = MP.
By the rule of signs, OQ = OP, ON = OM and NQ = -MP.
Now,
sin(360° - A) = \(\frac{NQ}{OQ}\)= \(\frac{-MP}{OP}\) = -sin A
cos(360° - A)= \(\frac{ON}{OQ}\)= \(\frac{OM}{OP}\) = cos A
tan(360° - A) = \(\frac{NQ}{ON}\)= \(\frac{-MP}{OM}\) = -tan A
cosec(360° - A)= \(\frac{OQ}{NQ}\)= \(\frac{OP}{-MP}\) = -cosec A
sec(360° - A)= \(\frac{OQ}{ON}\)= \(\frac{OP}{OM}\) = sec A
cot(360° - A)= \(\frac{ON}{NQ}\)= \(\frac{OM}{-MP}\) = -cot A
Ratios of negative angle (-A)
Let a revolving line OP start from OX and trace out an angle XOP = A. Let another line OQ = (-OP) revolve from OX to OX and return back to OQ through an angle XOQ = A.
Then \(\angle\)XOQ = -A.
Draw perpendiculars PM and QN from P and Q respectively to OX. Then similarly, as above,
OQ = OP, ON = OM and NQ = -MP
Then,
sin(-A) = \(\frac{NQ}{OQ}\)= \(\frac{-MP}{OP}\) = -sin A
cos(-A)= \(\frac{ON}{OQ}\)= \(\frac{OM}{OP}\) = cos A
tan(-A) = \(\frac{NQ}{ON}\)= \(\frac{-MP}{OM}\) = -tan A
cosec(-A)= \(\frac{OQ}{NQ}\)= \(\frac{OP}{-MP}\) = -cosec A
sec(-A)= \(\frac{OQ}{ON}\)= \(\frac{OP}{OM}\) = sec A
cot(-A)= \(\frac{ON}{NQ}\)= \(\frac{OM}{-MP}\) = -cot A
Values of every trigonometric ratio in different quadrants are:
1st quadrant | 2nd quadrant | 3rd quadrant | 4th quadrant |
all the ratios are (+e) positive | sin and cosec (+ve) remaining (-ve) | tan and cot (+ve) remaining (-ve) | cos and sec (+ve) remaining (-v) |
LHS =
= sin 9° × sin 27° = sin(90° - 81°) sin(90° - 63°)
= cos 63° × cos 81° = RHS
\(\therefore\) LHS = RHS proved.
Soln
LHS = sin1°. sin2°. sin45° .sec88° .sec89° = 1
= sin1°.sin2°.1.sec(90° - 2°).sec(90° - 1°)
= sin1°.sin2°.1.cosec2°.cosec1°
= sin1°.cosec1°.sin2°.cosec2°
= 1×1
= 1 = RHS
\(\therefore\) LHS = RHS
LHS = sec\(\theta\).cosec\(\theta\)(90° - \(\theta\)) - tan\(\theta\).cot(90° - \(\theta\)) = 1
= sec\(\theta\).sec\(\theta\) - tan\(\theta\).tan\(\theta\)
= sec^{2}\(\theta\) - tan^{2}\(\theta\)
= 1
= 1 = RHS
\(\therefore\) LHS = RHS proved.
Soln
LHS = \(\frac{cos(90°+θ).sec(-θ).tan(180° - θ)}{sec(360° + θ).sin(180°+θ).cot(90°-θ)}\) = -1
= \(\frac{cos(90°+θ).sec(-θ).tan(180° - θ)}{sec(4×90° + θ).sin(2×90°+θ).cot(90°-θ)}\)
= \(\frac{-sinθ.secθ(-tanθ)}{secθ(-sinθ).tanθ}\)
= -1 = RHS
\(\therefore\) LHS = RHS proved.
Soln
LHS = sin70°.cos20° + sin20°.cos70°
= sin70°.cos(90° - 70°) + sin(90° - 70°).cos70°
= sin70°. sin70° + cos70°.cos70°
= sin^{2}70° + cos^{2}70° = 1
If sin 9° × sin 27° = cos 63° × cos 81° then, what's it identity value?
What is the value of this identity sin1°. sin2°. sin45° .sec88° .sec89° ?
sec( heta).cosec(90° - ( heta)) - tan( heta).cot(90° - ( heta))
(frac{cos(90°+θ).sec(-θ).tan(180° - θ)}{sec(360° + θ).sin(180°+θ).cot(90°-θ)})
What is the numerical value of sin70°.cos20° + sin20°.cos70°?
What is the identical value of sin 240°?
What is the numerical value of tan(-300°)?
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Krischal Dhungel
Find the value of x: cosec(90 a) xcosa.cos(90 a)=sin(90 a)
Mar 10, 2017
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9999
prove that: {sin(-a)tan(180 a)sin(180 a)sec(270 a)}/{sin(360-a)cos(270-a)cosec(180-a)cot(360-a)}=-tan^2a
Feb 11, 2017
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