Question

show that 1/ cos theta - cos theta= tan theta - sin theta​

Answer 1

Step-by-step explanation:

SWER

Consider the L.H.S

sinθ+cosθ−1sinθ−cosθ+1

=(sinθ+cosθ−1sinθ−cosθ+1)×(sinθ+cosθ+1sinθ+cosθ+1)

=(sinθ+cosθ−1sinθ+1−cosθ)×(sinθ+cosθ+1sinθ+1+cosθ)

=(sinθ+cosθ)2−12(sinθ+1)2−cos2θ

=sin2θ+cos2θ+2sinθcosθ−1sin2θ+1+2sinθ−cos2θ

 

Since, sin2θ+cos2θ=1

 

Therefore,

=1+2sinθcosθ−11−cos2θ+1+2sinθ

Answer 2

$\Large{\underline{\underline{\bf{Correct Question:-}}}}$

❥ 1 / cos θ - cos θ = tan θ . sin θ

$\Large{\underline{\underline{\bf{AnSweR:-}}}}$

❥ 1 / cos θ - cos θ = tan θ . sin θ

$\Large{\underline{\underline{\bf{SoLuTion:-}}}}$

❥ We have to show that, 1 / cos θ - cos θ = tan θ . sin θ.

❧ We can prove this equation by considering LHS of the equation.

❥ LHS = 1 / cos θ - cos θ

❥ LHS = ( 1 - cos² θ ) / cos θ

❥ LHS = sin² θ / cos θ - - [ ∵ sin² θ + cos² θ = 1 ]

❥ LHS = sin θ × sin θ / cos θ

❥ LHS = sin θ / cos θ × sin θ

❥ LHS = tan θ . sin θ - - [ ∵ tan θ = sin θ / cos θ ]

∴ LHS = RHS

ྉ Hence proved

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❧ We can prove the given equation by considering RHS of the equation.

❥ RHS = tan θ . sin θ

❥ RHS = sin θ / cos θ × sin θ - - [ ∵ tan θ = sin θ / cos θ ]

❥ RHS = sin θ × sin θ / cos θ

❥ RHS = sin² θ / cos θ

❥ RHS = ( 1 - cos² θ ) / cos θ - - [ ∵ sin² θ + cos² θ = 1 ]

❥ RHS = 1 / cos θ - cos² θ / cos θ - - [ Separating the denominator ]

❥ RHS = 1 / cos θ - cos θ

∴ RHS = LHS

Hence proved

▩━━━━━━◈━━━━━━▩

$\Large{\underline{\underline{\bf{Additional Information:-}}}}$

ྉ Trigonometric Identities:

1. tan θ = sin θ / cos θ

2. sin² θ + cos² θ = 1

3. 1 + tan² θ = sec² θ

4. 1 + cot² θ = cosec² θ

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