Question

Evaluate the following :
(i) (1 + tan θ + sec θ) (1 + cotθ − cosec θ)
(ii) (sin θ + cos θ)² + (sin θ − cos θ)²
(iii) (sec²θ − 1) (cosec²θ −1)

Answer 1

(i) (1 + tan θ + sec θ) (1 + cotθ − cosec θ)
= 1(1 + cotθ - cosecθ) + tanθ(1 + cotθ - cosecθ) + secθ(1 + cotθ - cosecθ)
= 1 + cotθ - cosecθ + tanθ + tanθ.cotθ - tanθ.cosecθ + secθ + secθ.cotθ - cosecθ.secθ

we know, tanθ.cotθ = 1
tanθ.cosecθ = secθ
secθ.cotθ = cosecθ
cosecθ. secθ = tanθ + cotθ

= 1 + cotθ - cosecθ + tanθ + 1 - secθ + secθ + cosecθ - tanθ - cotθ
= 2

(ii) (sin θ + cos θ)² + (sin θ − cos θ)²
we know,
(sinθ + cosθ)² = 1 + sin2θ
(sinθ - cosθ)² = 1 - sin2θ

= 1 + sin2θ + 1 - sin2θ
= 2

(iii) (sec²θ - 1)(cosec²θ - 1)

we know, sec²θ - 1 = tan²θ
cosec²θ - 1 = cot²θ
(sec²θ - 1)(cosec²θ - 1) = tan²θ.cot²θ
= (tanθ.cotθ)² = 1² = 1

Answer 2

Hi ,

Here I'm using A instead of theta .

i ) ( 1 + tanA + secA )( 1 + cotA -cosecA )

= [1+sinA/cosA+1/cosA][1+cosA/sinA-1/sinA]

= [(cosA+sinA+1)/cosA][(sinA+cosA-1)/sinA]

= [(sinA + cosA)² - 1² ]/( sinAcosA )

= [ (sin²A+cos²A+2sinAcosA)-1]/(sinA cosA)

= ( 1 + 2sinAcosA - 1 )/( sinAcosA )

= ( 2sinAcosA )/( sinAcosA )

= 2

ii ) ( sinA + cosA )² + ( sinA - cosA )²

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( a + b )² + ( a - b )² = 2( a² + b² )

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Here ,

a = sinA , b = cosA

= 2 [ sin² A + cos² A ]

= 2 ( Since , sin² A + cos² A = 1 ]

iii )

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we know the trigonometric identity:

sec² A - 1 = tan² A ,

cosec²A - 1 = cot² A

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Now ,

( sec² A - 1 )( cosec²A - 1 )

= tan² A × cot² A

= 1 [ Since , tanAcotA = 1 ]

I hope this helps you.

: )