## Prove that secq(1-sinq)(secq + tanq) = 1

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

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

## Prove that (Sinq + Cosecq)2 + (Cosq + Secq)2 = 7 + Tan2q + Cot2q

we have to prove (Sinq + Cosecq)2 + (Cosq + Secq)2 = 7 + Tan2q + Cot2q we will start with LHS (Left hand side) and prove that it is equal to 7 + Tan2q + Cot2q Hence we will get LHS = RHS

Proof

LHS = (sinq + cosecq)2 + (cosq + secq)2
= (sin2q + cosec2q + 2 sinq cosecq ) + ( cos2q + sec2q + 2 cosq secq )
= sin2q + cosec2q + cos2q + sec2q + 2 + 2 Hint :2 * sinq * 1/sinq + 2 * cosq * 1/cosq

= ( sin2q + cos2q) + cosec2q + sec2q + 2 + 2

= 1 +cosec2q + sec2q + 4

= (1 + cot2q) + (1 + tan2q) + 5

= 7 + tan2q + cot2q
= RHS
Hence proved RHS = LHS ie (Sinq + Cosecq)2 + (Cosq + Secq)2 = 7 + Tan2q + Cot2q

See in Picture

## cpct full form in maths

cpct stands for Corresponding parts of congruent triangles, if two triangle are congruent their corresponding sides are equal also their corresponding angles are equal

Suppose below triangle 1 and 2 are congruent

Then their (cpct) Corresponding parts of congruent triangles are also equal means

AB=XY
AC=XZ
BC=YZ

## Prove that tanq+Secq-1÷tanq-secq+1=cosq÷1-sinq

Here am going to explain the steps Prove that tanq+Secq-1÷tanq-secq+1=cosq÷1-sinq

LHS => tanq+Secq-1/tanq-secq+1

=> (sinQ/cosQ+1/cosQ-1)/(sinQ/cosQ-1/cosQ+1) (divide nominator and denominate by cosQ)

=> (sinQ+1-cosQ)/(sinQ-1+cosQ) (Multipy nominator and denominate by (1-sinQ))

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

=> (sinQ+1-cosQ-sinQ2-sinQ+sinQ*cosQ)/(1-sinQ)(sinQ-1+cosQ)

=> (1-cosQ-sinQ2+sinQ*cosQ)/(1-sinQ)(sinQ-1+cosQ)
=> (1-sinQ2-cosQ+sinQ*cosQ)/(1-sinQ)(sinQ-1+cosQ)

=> (cosQ2-cosQ+sinQ*cosQ)/(1-sinQ)(sinQ-1+cosQ)
=> cosQ(cosQ-1+sinQ)/(1-sinQ)(sinQ-1+cosQ)

=> cosQ(cosQ-1+sinQ)/(1-sinQ)(cosQ-1+sinQ)
=> cosQ/(1-sinQ) = RHS

Hence proved