Question 25534: 1)prove that cosA + sinA tanA = secA
2)prove that cos(180-y) = -cos y
3)use the sum and difference identities to find the exact value of tan 165 degrees
Answer by venugopalramana(3286)   (Show Source):
You can put this solution on YOUR website! )prove that cosA + sinA tanA = secA
LHS=COS(A)+SIN A * SIN(A) / COS(A)= {COS(A) * COS(A) + SIN(A)*SIN(A)} / COS(A)
=1/COS(A)=SEC(A) ..SINCE {COS (A)}^2 +{(SIN (A)}^2 = 1
2)prove that cos(180-y) = -cos y
GO BACK TO THE DEFINITION OF TRIGNOMETRIC RATIOS.
WE START WITH THE ORIGIN AND THE 2 AXES X'OX AND Y'OY
WE START WITH A LINE SEGMENT OP ON X AXIS CALLED RADIUS VECTOR.BY CONVENTION THIS IS ALWAYS POSITIVE.WE LET IT TRAVEL IN COUNTERCLOCKWISE DIRECTION (CCD IN SHORT)WHICH IS CONSIDERED THE POSITIVE DIRECTION (PD IN SHORT ), MAKING AN ANGLE ..SAY...A... WITH OX IN THE CCD OR PD.DEPENDING ON THE ANGLE TRAVELLED IT MAY END UP IN ANY OF THE 4 QUADRANTS....YOX (I QUADRANT...0 TO 90 DEGREES),YOX'(II QUADRANT...90 TO 180 DEGREES),Y'OX'(III QUADRANT....180 TO 270 DEGREES),Y'OX (IV QUADRANT...270 TO 360 DEGREES).
THEN WE DROP A PERPENDICULAR PQ FROM P TO X'OX.WE CONSIDER THE RIGHT ANGLED TRIANGLE OPQ TO DEFINE THE TRIGNOMETRIC RATIOS AS FOLLOWS, WITH ANGLE AT O THAT IS ANGLE POQ BEING A
SIN A = QP/OP
COS A =OQ /OP
TAN A =QP/OQ
AS GIVEN IN THE BEGINING OP IS ALWAYS POSITIVE .SO SIGN OF A RATIO DEPENDS ONLY ON SIGN OF OQ AND QP.
NOW IN YOUR PROBLEM
COS (180-Y)...OP STARTED ON X'OX AND TRAVELLED 180 DEGREES IN CCD ENDING UP ON OX'.THEN IT TRAVELLED BACK BY Y DEGREES IN THE NEGATIVE OR CLOCK WISE DIRECTION TO END UP IN II QUADRANT.
SO IN OUR CASE IN THE RIGHT ANGLED TRIANGLE OPQ ANGLE POQ = Y....OQ IS NEGATIVE SINCE AT THIS POSITION OQ IS TOWARDS OX' THE NEGATIVE X AXIS.
HENCE COS (180-Y)= OQ / OP = - COS Y.
3)use the sum and difference identities to find the exact value of tan 165 degrees
TAN 165 = TAN (180-15)= - TAN 15 = - TAN (45-30)
= - {(TAN 45 - TAN 30 )/ ( 1 + TAN 45 * TAN 30 ) }
= - { (1-1/SQ.RT.3)/ ( 1+1*1/SQ.RT.3)}
= - { (SQ.RT.3-1)/(SQ.RT.3 +1)}= - { ( SQRT.3 - 1 )^2/((SQRT.3+1)*(SQRT.3-1))}
= - (3+1-2*SQRT.3)/(3-1)= - (2-SQRT.3)=SQRT.3-2
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