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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

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

Here we have to prove secq(1-sinq)(secq + tanq) = 1 we will start with LHS (Left hand side ie secq(1-sinq)(secq + tanq) ) and prove that it is equal to 1 Hence we will get LHS = RHS
Our aim is to Show that secq(1-sinq)(secq + tanq) = 1

Proof

Lets start with LHS of the equations
LHS = secq(1-sinq) (secq + tanq)

= (secq-sinq×secq) (secq+tanq)

= (secq-tanq) (secq+tanq) hint : secq=1/cosq and sinq/cosq=tanq

= (sec²q-tan²q) hint : (x+y)(x-y)=x²-y²

= 1 hint : we know sec²q-tan²q =1
= RHS
Hence proved RHS = LHS ie secq(1-sinq)(secq + tanq) = 1