Question

(8) seco + tano =
cos e/
1-sin e​

Answer 1

Answer:

You use the trigonometric identity

Step-by-step explanation:

sece+ tan e =LHS

so, 1/cos e + sin e/ cos e

=1+sin e/cos e

squaring and taking roots

root of (1 + sin e) squared/cos squared e

= root of (1+ sine)(1+ sine e)/1- sin squared e

... sin squared e + cos squared e = 1

and (a square - b square) = ( a+b) (a-b)

so root of (1+ sin e)( 1+ sin e)/( 1+ sin e)( 1- sin e)

= root of ( 1+ sin e)/ (1 - sin e)=LHS

RHS= cos e/1- sine

taking roots and squares again,

root of cos squared e /(1- sin e) squared

= (1 -sin e)(1 + sin e) /( 1 - sine)(1 - sin e)

= 1 +sin e / 1 -sin e

so, LHS = 1+ sine/ 1- sin e

RHS = 1 + sin e/ 1- sin e

HENCE, LHS =RHS

Proved