Prove that (cotQ+ cosecQ -1) / (cotQ – cosecQ +1) = (1-cosQ) / sinQ?

Here we have to prove (cotQ+ cosecQ -1) / (cotQ – cosecQ +1) = (1-cosQ) / sinQ we will start with LHS (Left hand side ie (cotQ+ cosecQ -1) / (cotQ – cosecQ +1) and prove that it is equal to (1-cosQ) / sinQ Hence we will get LHS = RHS

Proof

Lets start with LHS of the equations
LHS = (cotQ+ cosecQ -1) / (cotQ – cosecQ +1)

= (cosQ/ sinQ + 1/sinQ – sinQ / sinQ) / (cosQ / sinQ – 1 / sinQ + sinQ / sinQ)

= (cosQ+ 1 – sinQ) / (cosQ – 1 + sinQ) * (cosQ + 1 – sinQ) / (cosQ + 1 – sinQ)

= (cosQ + 1 – sinQ) (cosQ + 1 – sinQ) / ( ((cosQ – 1 + sinQ) (cosQ + 1 – sinQ))

= (cosQ + 1 – sinQ)² / (cos²Q – (1 – sinQ)²)

= (cos²Q – 2 cosQ (1 – sinQ) + (1 – sinQ)²) / (cos²Q – 1 + 2 sinQ – sin²Q)

= (cos²Q – 2 cosQ + 2 sinQ cosQ + 1 – 2 sinQ + sin²Q) / (1 – sin² Q – 1 + 2 sinQ – sin²Q)
= (2 – 2 cosQ + 2 sinQ cosQ – 2 sinQ) / (2 sinQ – 2 sin²Q)
= 2 (1 – cosQ) (1 – sinQ) / (2 sinQ (1 – sinQ) )
= 1 – cosQ / sinQ
= RHS

Prove that sinQ cotQ + sinQ cosecQ = 1 + cosQ

Here we have to prove sinQ cotQ + sinQ cosecQ = 1 + cosQ we will start with LHS (Left hand side ie sinQ cotQ + sinQ cosecQ ) and prove that it is equal to 1 + cosQ Hence we will get LHS = RHS

Proof

Lets start with LHS of the equations
LHS = sinQ cotQ + sinQ cosecQ
= sinQ (cosQ/sinQ) + sinQ * 1/sinQ

= cosQ + 1
= 1 + cosQ

Prove that sinQ/cotQ + cosecQ = 2 + sinQ/cotQ – cosecQ

Here we have to prove sinQ/cotQ + cosecQ = 2 + sinQ/cotQ – cosecQ we will start with LHS (Left hand side ie sinQ/cotQ + cosecQ ) and prove that it is equal to 2 + sinQ/cotQ – cosecQ Hence we will get LHS = RHS

Our aim is to Show that sinQ/cotQ + cosecQ = 2 + sinQ/cotQ – cosecQ

Proof

Lets start with LHS of the equations
LHS = sinQ/(cotQ + cosecQ)

= sinQ/(cosQ/sinQ + 1/sinQ)

= sinQ2 /(cosq+1)/sinQ

= sinQ2 /(1 + cosQ)

= sin²Q (1-cosQ) /(1 + cosQ)(1 – cosQ)
= sin²Q (1-cosQ) /(1 – cos2Q)
= sin²Q (1-cosQ) /sin²Q Hint : (1 – cos²Q) = sin²Q since sin²Q + cos²Q = 1
= 1-cosQ

RHS = 2 + sinQ/cotQ – cosecQ
= 2+ sinQ/(cosQ/sinQ-1/sinq)
= 2 +sin²q/(cosQ – 1)
= 2 -sin²q/(1-cosQ )
= 2 – sin²q(1 + cosQ)/(1-cosQ) (1 + cosQ)
= 2 – sin²q(1 + cosQ)/(1-cos²Q)
= 2 – sin²q(1 + cosQ)/sin²q Hint : (1 – cos²Q) = sin²Q since sin²Q + cos²Q = 1
= 2 – (1 + cosQ)
= 1 – cosQ
RHS = LHS

Hence proved

Prove that sinQ/cotQ + cosecQ = 2 + sinQ/cotQ – cosecQ

Basic Trigonometric Formulas

based on right-angled triangle their 6 ratios in Trigonometry, this are basically called Trigonometric functions .The six trigonometric functions are sine, cosine, secant, co-secant, tangent and co-tangent.The value of each function can be find out using below trigonometric formulas

Here are the definition of each of this trigonometric formulas

trigonometric formulas

sin θ = opposite Side/hypotenuse where θ is the angle cos θ = adjacent Side/hypotenuse where θ is the angle tan θ = opposite Side/adjacent Side where θ is the angle sec θ = hypotenuse/adjacent Side where θ is the angle cosec θ = hypotenuse/opposite Side where θ is the angle cot θ = adjacent Side/opposite Side where θ is the angle

As per the above table definition we can relate each of the formula in below ways

cosec θ = 1/sin θ    hint :  1/opposite Side/hypotenuse  = hypotenuse/opposite

sec θ = 1/cos θ      hint :  1/adjacent Side/hypotenuse   = hypotenuse/adjacent Side

cot θ = 1/tan θ      hint :  1/opposite Side/adjacent   = adjacent Side/opposite Side

sin θ = 1/cosec θ    hint :  1/hypotenuse/opposite Side   = opposite Side/hypotenuse

cos θ = 1/sec θ      hint :  1/hypotenuse/adjacent Side   =adjacent Side/hypotenuse 

tan θ = 1/cot θ      hint :  1/adjacent Side/opposite Side   = opposite Side/adjacent Side 


we can find each of the six trigonometric functions values using above trigonometric formulas