Math, asked by vamikabahety98, 10 months ago

29. Pjove that: sec 0 + tan 0-1
tan - sec 0 +1
1+ sin
cos 0​

Answers

Answered by Rapanzeel
2

Answer:

According to given sum,

sin (1+tan)+cos (1+cot)=sec+cosec

=> LHS= sin (1+tan)+cos (1+1/tan)

=> sin (1+tan)+cos (1+tan)/tan

=> (1+tan)(sin+cos/tan)

=> (1+tan)(sin.tan+cos)/tan

=> (1+tan)(sin^2/cos+cos)/tan

=> (1+tan)(sin^2+cos^2)/tan.cos

=>(1+tan)/tan.cos

=>(1/tan.cos)+tan/tan.cos

=>( cot/cos)+(1/cos)

=> cosec +sec

LHS= RHS.

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Answered by rajanisingha
2

According to given sum,

sin (1+tan)+cos (1+cot)=sec+cosec

=> LHS= sin (1+tan)+cos (1+1/tan)

=> sin (1+tan)+cos (1+tan)/tan

=> (1+tan)(sin+cos/tan)

=> (1+tan)(sin.tan+cos)/tan

=> (1+tan)(sin^2/cos+cos)/tan

=> (1+tan)(sin^2+cos^2)/tan.cos

=>(1+tan)/tan.cos

=>(1/tan.cos)+tan/tan.cos

=>( cot/cos)+(1/cos)

=> cosec +sec

LHS= RHS.

Like my answer if you find it useful!

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