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tanA + secA -1 / tanA -secA + 1 = 1 +sinA / cosA
I would like to get answer for this question in trigonometry

n , 7 Years ago
Grade 10
anser 2 Answers
Nandana

Last Activity: 7 Years ago

hi ,
 tanA + secA -1 / tanA -secA + 1 = (sinA +1-cosA) /(sinA+cosA-1 ) × (sinA+cosA +1)/(sinA+cosA+1)
                        = (sinA+cosA )(sinA-cosA)+sinA+cosA-cosA+1+sinA-cosA/(sinA+cosA)2-12)
                                  after simplification this is equal to
                        = 2sinA(1+sin A)/ 2sinA cosA
                        = (1+sin A)/  cosA
                         hence  right side obtainnd
                          

Kushagra Madhukar

Last Activity: 4 Years ago

Dear student,
Please find the answer to your question below.
 
tanA + secA – 1 / tanA – secA + 1 = (sinA + 1 – cosA) / (sinA + cosA – 1) × (sinA + cosA + 1) / (sinA + cosA + 1)
                        = (sinA + cosA )(sinA – cosA) + sinA + cosA – cosA + 1 + sinA – cosA / (sinA + cosA)2 – 12)
On simplifying, we get,
                        = 2sinA(1 + sin A) / 2sinAcosA
                        = (1 + sin A) /  cosA
LHS = RHS
Hence proved.
 
Hope it helps.
Thanks and regards,
Kushagra

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