Question

tan^2theta +sec^2theta plus 1=2+2cot^2theta by cosec^2theta-1​

Answer 1

Question : -

tan² @ + sec² @ + 1 = (2 + 2cot² @)/(cosec² @ - 1)

ANSWER

Given : -

Prove that ;

tan² @ + sec² @ + 1 = (2 + 2cot² @)/(cosec² @ - 1)

Required to prove : -

• LHS = RHS

Identity used : -

• sin² @ + cos² @ = 1

Solution : -

tan² @ + sec² @ + 1 = (2 + 2cot² @)/(cosec² @ - 1)

Consider the LHS part;

tan² @ + sec² @ + 1

Here, let's simplify this LHS part

➔ tan² @ + sec² @ + 1

➔ (sin² @)/(cos² @) + (1)/(cos² @) + 1

Taking cos² @ as LCM

➔ (sin² @ + 1 + cos² @)/(cos² @)

➔ ([sin² @ + cos² @ ]+ 1)/(cos² @)

since, sin² @ + cos² @ = 1

➔ (1+1)/(cos² @)

➔ (2)/(cos² @)

Now,

Let's consider the RHS part;

(2 + 2cot² @)/(cosec² @ - 1)

Here, let's simplify this further

➔ (2 + 2cot² @)/(cosec² @ - 1)

➔ (2 + 2 x [cos² @]/[sin² @] )/( [1]/[sin² @] - 1)

➔ (2 + [2cos² @]/[sin² @])/([1]/[sin² @] - 1)

➔ ([2sin² @ + 2cos² @]/[sin² @])/([1 - sin² @]/[sin² @])

sin² @ get's cancelled on both numerator & denominator

➔ (2sin² @ + 2cos² @)/(1 - sin² @)

Taking 2 as common in numerator

➔ (2[sin² @ + cos² @])/(1 - sin² @)

since, sin² @ + cos² = 1

➔ (2[1])/(cos² @)

➔ (2)/(cos² @)

LHS = RHS

➔ Hence Proved !