Trigonometry> Prove that secA+ tanA-1÷ tanA- secA+1= co...

Prove that secA+ tanA-1÷ tanA- secA+1= cosA÷1- sinA

gokul , 8 Years ago

Grade 10

3 Answers

Yashaswi Agrawal

Last Activity: 8 Years ago

= 1+sinA

cosA

= 1+sinAx(1-sinA)

cosAx(1-sinA)

= 1-sin²A

cosA(1-sinA)

= cos²A

cosA(1-sinA)

= cosA

1-sinA

Hence Proved

KaranKishoreDv

Last Activity: 7 Years ago

Multiply ((tanA-1)+(secA))/ ((tanA+1)-secA)) by ((tanA-1)+(secA)) on numerator and denominator. Then use the identitype a²-b²to proceed further. Thanks.

Rishi Sharma

Last Activity: 5 Years ago

Dear Student,
Please find below the solution to your problem.

secA+tanA-1)/tanA-secA+1)
We know that sec^2-tan^2=1
=(secA+tanA-(sec^2 A - tan ^2A)) /tanA-secA+1)
We also know that a^2-b^2=(a+b)(a-b)
= (sec A + tan A - (sec A + tan A) ( sec A - tan A)) / ( tanA-secA+1)
= ( sec A + tan A ) ( 1- (sec A - tan A)/ ( tanA-secA+1)
= (sec A + tan A)/(+ tan A - sec A)/(tan A- sec A+ 1)
= (sec A + tan A)
= 1/cos A + sin A/ cos A
= (1+ sin A)/ cos A
= (1 + sin A )(1- sin A)/(cos A (1- sin A))
= (1- sin ^2 A/(cos A (1- sin A))
= cos ^2 A / (cos A (1- sin A))
= cos A /(1- sin A)

Thanks and Regards