Question

tan2=1-a2

proove sec +tan3cosec=(2-a2)3 / 2

Answer 1

QUESTION :

tan2=1-a2

proove sec +tan3cosec=(2-a2)3 / 2

CORRECTION IN QUESTION :

If tan²A = 1-a²

then prove that

secA+tan³A.cosecA = (2-a²)^(3/2)

ANSWER :

GIVEN :

tan²A = 1-a²

TO PROVE :

secA+tan³A.cosecA = (2-a²)^(3/2)

PROOF :

now,

LHS=

secA+tan³A.cosecA

so by multiplying and dividing numerator and denominator by secA

we get,

secA+tan³A.cosecA

= secA(secA+tan³A.cosecA)/secA

now by separating the terms

= secA((secA/secA)+(tan³A.cosecA/secA))

now as,

cosA=1/secA

and

sinA=1/cosecA

we get

=secA(1+(tan³A.cosA/sinA))

now as,

cosA/sinA=cotA

we get

= secA(1+(tan³A.cotA))

also

1/tanA = cotA

we get

=secA(1+(tan³A/tanA))

=secA(1+tan²A)..........1

now

1+tan²A=sec²A .... identity

taking square root on both the sides

we get

secA=√(1+tan²A).....2

so put 2 in 1

secA(1+tan²A)

= (√(1+tan²A))(1+tan²A)

= (1+tan²A)^(1/2) (1+tan²A)¹

so by using expotential identity

aⁿ × a^m

= (a)^(m+n)

so we get

= (1+ tan²A)^((1/2)+1)

= (1+tan²A)^((1+2)/2)

= (1+tan²A)^(3/2)

now but it is given that

tan²A=1-a²

so we get

= (1+(1-a²))^(3/2)

=(2-a²)^(3/2)

=RHS

hence LHS=RHS

hence proved

NOTE :

while solving such questions try to simplify the given equation

also as we had a given value in tanA try to convert other trigonometric ratios in tan form to so it will be easy to get the answer