You visited us 1 times! Enjoying our articles? Unlock Full Access!

Trigonometric Identities

prove that in...

Question

prove that inA-cosA+1÷sinA+cosA-1=1÷secA-tanA using identity sec2A = 1+tan2A

Open in App

Solution

(sin A-cos A+1)/(sin A+cosA-1)=1/(sec A-tan A)

L.H.S. divide above and below by cos A

=(tan A-1+secA)/(tan A+1-sec A)

=(tan A-1+secA)/(1-sec A+tan A)

We know that 1+tan^2(A)=sec ^2(A)

Or put in deminator 1=sec^2(A)-tan ^2(A)=(sec A+tan A)(secA-tanA)

=(sec A+tan A-1)/[(sec A+tan A)(sec A-tan A)-(sec A-tan A)]

=(sec A+tan A-1)/(sec A-tan A)(sec A+tan A-1)

= 1/(sec A-tan A) , proved.
LHS =RHS

Suggest Corrections

Similar questions

secA - t a n A = \frac{cos A}{1 + sin A}

\frac{t a n A + s e c A - 1}{t a n A - s e c A + 1} = \frac{1 + s i n A}{c o s A}

{(cosec θ - cot θ)}^{2} = \frac{1 - cos θ}{1 + cos θ}

\frac{cos A}{1 + sin A} + \frac{1 + sin A}{cos A} = 2 sec A

\frac{tan θ}{1 - cot θ} + \frac{cot θ}{1 - tan θ} = 1 + sec θ cosec θ

sin θ

cos θ

\frac{1 + sec A}{sec A} = \frac{{sin}^{2} A}{1 - cos A}

\frac{cos A - sin A - 1}{cos A + sin A + 1} = cosec A + cot A

{cosec}^{2} A = 1 + {cot}^{2} A

\sqrt{\frac{1 + sin A}{1 - sin A}} = sec A + tan A

\frac{sin θ - 2 {sin}^{3} θ}{2 {cos}^{3} θ - cos θ}

{(sin A + cosec A)}^{2} + {(cos A + sec A)}^{2} = 7 + {tan}^{2} A + {cot}^{2} A

(cosec A - sin A) (sec A - cos A) = \frac{1}{tan A + cot A}

(\frac{1 + {tan}^{2} A}{1 + {cot}^{2} A}) = {(\frac{1 - tan A}{1 - cot A})}^{2} = {tan}^{2} A

Join BYJU'S Learning Program