Cos A - Sin A + 1/cos A+ sinA- 1 = 1+cosA/sinA
Answers
Answer:
To Prove : [ { sin³A + cos³A } / { sinA + cosA } ] + cosA sinA = 1
Proof :
From the properties of expansion we know that a³ + b³, when factorised, is ( a + b)( a² + b² - ab) .
Therefore,
sin³A + cos³A will be ( sinA + cosA )( cos²A + sin²A - cosAsinA )
Thus,
= > [ { ( sinA + cosA )( sin²A + cos²A - sinAcosA) } / { sinA + cosA } ] + sinAcosA
Removing ( sinA + cosA) from numerator and denominator both.
From the properties of trigonometry, we know : sin²A + cos²A
= > [ 1 - sinA cosA ] + sinAcosA
= > 1 - sinAcosA + sinAcosA
= > 1
Hence proved.
Solution :-
→ cosA-sinA+1/cosA+sinA-1
→ (cosA-sinA+1)(cosA+sinA+1)/(cosA+sinA-1)(cosA+sinA+1)
→ (cos²A-cosAsinA+cosA+cosAsinA-sin²A+sinA+cosA-sinA+1)/{(cosA+sinA)²-(1)²}
→ (cos²A-sin²A+2cosA+1)/(cos²A+2cosAsinA+sin²A-1)
→ {cos²A+2cosA+(1-sin²A)}/(1+2cosAsinA-1) [∵, sin²A+cos²A=1]
→ (cos²A+2cosA+cos²A)/2cosAsinA
→ (2cos²A+2cosA)/2cosAsinA
→ 2cosA(cosA+1)/2cosAsinA
→ (cosA+1)/sinA
→ cosA/sinA+1/sinA
→ cotA+cosecA
→ cosecA+cotA
Hence Proved