Question

prove that cos square theta + tan square theta minus one upon sin square theta is equals to tan square theta

Answer 1

COS²Q+TAN²Q-1 / SIN²Q = TAN²Q

L.H.S

==> COS²Q+TAN²Q-1 / SIN²Q

PUTTING FORMULA OF COS²Q+SIN²Q = 1 IN 1

==> COS²Q+TAN²Q-(SIN²Q+COS²Q) / SIN²Q

==> COS²Q + TAN²Q - SIN²Q - COS²Q ÷ SIN²Q

==> TAN²Q - SIN²Q ÷ SIN²Q

==> SIN²Q/COS²Q - SIN²Q ÷ SIN²Q

==> SIN²Q-SIN²Q*COS²Q / COS²Q ÷ SIN²Q

==> SIN²Q(1-COS²Q)/COS²Q ÷ SIN²Q

==>  SIN²Q(1-COS²Q)×1/COS²Q × 1/SIN²Q

==>  (1-COS²Q)× 1/COS²Q

==> SIN²Q × 1/COS²Q

==>  SIN²Q/COS²Q

==> TAN²Q

SO, TAN²Q == R.H.S

HENCE PROVED

I HOPE ITS HELPFUL

Answer 2

Step-by-step explanation:

your answer is an attachment.