Math, asked by adityashukla52, 8 months ago

Cos A - Sin A + 1/cos A+ sinA- 1 = 1+cosA/sinA​

Answers

Answered by Anubhavdeb
3

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.

Answered by Anonymous
2

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

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