Math, asked by AkankshaTinni, 1 year ago

to prove: 1/(sec A-1) + 1/(sec A+1)= 2 cosec A cot A

Answers

Answered by mysticd
83
Solution :

LHS = 1/(secA -1) + 1/(secA+1)

= [secA+1+secA-1]/[(secA-1)(secA+1)]

= (2secA)/[sec²A-1]

=(2secA)/tan²A

= (2/cosA)/(sin²A/cos²A)

= ( 2cosA)/sin²A

= 2(1/sinA)(cosA/sinA)

= 2cosecAcotA

= RHS

••••

AkankshaTinni: thank you
Answered by siddhartharao77
98

Given : \frac{1}{secA - 1} + \frac{1}{secA + 1}

=> \frac{secA - 1 + secA + 1}{(secA - 1)(secA + 1)}

We know that (a + b)(a - b) = a^2 - b^2

=> \frac{2secA}{(sec^2A - 1)}

We know that sec^2θ - tan^2θ = 1 (or) sec^2θ - 1 = tan^2θ

=> \frac{2secA}{tan^2A}

=> \frac{2}{cosA} * \frac{cos^2A}{sin^2A}

=> 2 * \frac{1}{sinA} *\frac{cosA}{sinA}

2 cosecA cotA.


Hope it helps!


AkankshaTinni: thanks
siddhartharao77: Welcome!
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