मराठी

2 (Sin6 θ + Cos6 θ) − 3 (Sin4 θ + Cos4 θ) is Equal to

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प्रश्न

2 (sin6 θ + cos6 θ) − 3 (sin4 θ + cos4 θ) is equal to 

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उत्तर

The given expression is `2(sin^6θ+cos^6θ)-3(sin^4θ+cos^4θ)` 

Simplifying the given expression, we have

`2(sinθ+cos^6θ)-3(sin^4θ+cos^4θ)` 

= `2sin^6θ+2cos^6θ-3sin^4θ-3cos^4θ`

=`(2 sin^6 θ-3sin^4θ)+(2 cos^6-3 cos^4θ)`

=`sin^4θ(2sin^2θ-3)+cos^4θ(2 cos^2θ-3)`

`=sin^4θ{2(1-cos^2)-3}+cos^4θ{2(1-sin^2 θ)-3)` 

`= sin^4θ(2-2cos^2θ-3)+cos^4θ(2-2sin^2 θ-3) `

`=sin^4θ(-1-2cos^θ)+cos^4θ(1-2sin^2θ)` 

`= -sin^4θ-2 sin^4θ cos^2θ-cos^4θ-2cos^4 θ sin^2θ`

`=sin^4θ-cos^4θ-2 cos^4 θ sin^2θ-2 sin^4 θcos^2θ`

`=-sin^4θ-cos^4θ-2cos^2θ sin^2(cos^2+sin^2θ)`

`=-sin^4θ-cos^4θ-2cos^2θsin^2θ(1)`

`=-sin^4θ-cos^4θ-2cos^2sin^2θ`

`=(sin^4θ+cos^4 θ+2 cos^2 θ sin^2 θ)`

`=-{(sin^2θ)^2+(cos^2θ)^2+2 sin^2 θ cos^2θ}`

` =-(sin^2θ+cos^2θ)^2` 

`=-(1)^2`

`=-1`

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पाठ 11: Trigonometric Identities - Exercise 11.4 [पृष्ठ ५७]

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आर.डी. शर्मा Mathematics [English] Class 10
पाठ 11 Trigonometric Identities
Exercise 11.4 | Q 15 | पृष्ठ ५७

संबंधित प्रश्‍न

Prove the following identities:

`(i) (sinθ + cosecθ)^2 + (cosθ + secθ)^2 = 7 + tan^2 θ + cot^2 θ`

`(ii) (sinθ + secθ)^2 + (cosθ + cosecθ)^2 = (1 + secθ cosecθ)^2`

`(iii) sec^4 θ– sec^2 θ = tan^4 θ + tan^2 θ`


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`(i) 2 (sin^6 θ + cos^6 θ) –3(sin^4 θ + cos^4 θ) + 1 = 0`

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`tan theta - cot theta = (2 sin^2 theta - 1)/(sin theta cos theta)`


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(cosec θ − sec θ) (cot θ − tan θ) = (cosec θ + sec θ) ( sec θ cosec θ − 2)


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`sqrt((1 - cosA)/(1 + cosA)) = sinA/(1 + cosA)`


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