Question

cotsquare theta(sectheta-1/1(sintheta)+secsquare theta(sintheta-1/1+sectheta)=0 prove that ​

Answer 1

Solution

cot²∅(sec∅ - 1)/(1 + sin∅) + sec²∅(sin∅ - 1)/(1 + sec∅) = 0

Part 1 → cot²∅(1/cos∅ - 1)/(1 + sin∅)

→ cos²∅/sin²∅ × (1 - cos∅)/cos∅ × 1/(1 + sin∅)

→ cos∅(1 - cos∅)/sin²∅(1 + sin∅)

→ cos∅(1 - cos∅)/(1 - cos∅)(1 + cos∅)(1 + sin∅)

→ cos∅/(1 + cos∅)(1 + sin∅)

Part 2 → sec²∅(sin∅ - 1)/(1 + sec∅)

→ 1/cos²∅ × (sin∅ - 1) × cos∅/(1 + cos∅)

→ (sin∅ - 1)/cos(1 + cos∅)

→ Part 1 + Part 2

→ cos∅/(1 + cos∅)(1 + sin∅) + (sin∅ - 1)/cos∅(1 + cos∅)

→ (cos²∅ + sin²∅ - 1)/cos∅(1 + cos∅)(1 + sin∅)

→ (1 - 1)/cos∅(1 + cos∅)(1 + sin∅)

→ 0/cos∅(1 + cos∅)(1 + sin∅) = 0

Q.E.D

Answer 2

$\bold {\huge {Your ~answer :-}}$

$\bf\huge\boxed{LHS=RHS}$

--------------------------------------------------

EXPLANATION —

cot²θ(secθ - 1)/(1 + sinθ) + sec²θ(sinθ - 1)/(1 + secθ) = 0

➡What we will do we will break into two part for easy calculation and finally will add them.

PART 1

=> cot²θ(1/cosθ - 1)/(1 + sinθ)

=> cos²θ/sin²θ × (1 - cosθ)/cosθ × 1/(1 + sinθ)

=>cosθ(1 - cosθ)/sin²θ(1 + sinθ)

=> cosθ(1 - cosθ)/(1 - cosθ)(1 + cosθ)(1 + sinθ)

=>cosθ/(1 + cosθ)(1 + sinθ)

--------------------------------------------------

PART 2

=> sec²θ(sinθ - 1)/(1 + secθ)

=> 1/cos²θ × (sinθ - 1) × cosθ/(1 + cosθ)

=> (sinθ - 1)/cos(1 + cosθ)

--------------------------------------------------

Now,

Final result = PART 1 + PART 2

=> cosθ/(1 + cosθ)(1 + sinθ) + (sinθ - 1)/cosθ(1 + cosθ)

=> (cos²θ + sin²θ - 1)/cosθ(1 + cosθ)(1 + sinθ)

=>(1 - 1)/cosθ(1 + cosθ)(1 + sinθ)

=> 0/cosθ(1 + cosθ)(1 + sinθ) = 0

Hence it is proved

As LHS = RHS

$\huge{\red{\ddot{\smile}}}$