Question

CAN ANYONE SOLVE THIS BY RHS​

Answer 1

Step-by-step explanation:

Solution:-

We know that

Sin² θ+ Cos² θ = 1

=> Cos² θ = 1 - Sin² θ

=> Cos θ × Cos θ = (1+Sin θ)(1-Sin θ)

=> Cos θ/(1-Sin θ) = (1+Sin θ)/Cos θ

By the property of equal ratio

=> Cos θ/(1-Sin θ) = (1+Sin θ)/Cos θ =

(1+Sin θ-Cos θ) /( Cos θ- 1+ Sin θ)

=> Cosθ/(1-Sinθ) = (1+Sinθ-Cosθ)/(Cosθ- 1+Sinθ)

=> Cosθ/(1-Sinθ) = (Sinθ-Cosθ+1)/(Cosθ+Sinθ-1)

On dividing LHS by Cos θ

=> (Cos θ/Cos θ)/(1- Sin θ)/Cos θ) =

(Sinθ-Cosθ+1)/(Cosθ+Sinθ-1)

=> 1/(1- Sin θ)/Cos θ) = (Sinθ-Cosθ+1)/(Cosθ+Sinθ-1)

=> 1/(1/Cos θ)-(Sin θ/ Cos θ) =

(Sinθ-Cosθ+1)/(Cosθ+Sinθ-1)

=> 1/(Secθ-Tanθ) = (Sinθ-Cosθ+1)/(Cosθ+Sinθ-1)

=> RHS = LHS

Used formulae:-

• Tan A = Sin A / Cos A
• Sec A = 1/ Cos A
• Sin² A + Cos² A = 1
• Cos² A = 1 - Sin² A
• If a/b = c/d then a/b = c/d

= (a+c)/(b+d)

Answer 2

Step-by-step explanation:

Solution-

L.H.S = (sin θ - cos θ + 1)/(sin θ + cos θ - 1)

= (tan θ - 1 + sec θ)/(tan θ + 1 - sec θ)

= [(tan θ + sec θ) - 1]/[(tan θ - sec θ) + 1]

= [{(tan θ + sec θ) - 1} (tan θ - sec θ)]/[{(tan θ - sec θ) + 1}( tan θ - sec θ)]

= [(tan² θ - sec² θ) - (tan θ - sec θ)]/[(tan θ - sec θ + 1)(tan θ - sec θ]

= -1 - tan θ + sec θ/[(tan θ - sec θ + 1)(tan θ - sec θ)]

= -1/(tan θ - sec θ)

= 1/(sec θ - tan θ)

= R.H.S

Remember:- sec² θ = 1 + tan² θ.