Prove The Following - 1 (Trigonometry)

This note facilitates you to prove good Trigonometry questions. So please use this note. I will try my best to post these types of notes from now.


\(\color{green}\text{Prove the following Identities :}\)

  1. \(\large\frac{1 + sin A}{1 - sin A} = \frac{cosec A + 1}{cosec A - 1}\) \(\boxed{\text{proved in 1 method}}\)

  2. \(\frac{1}{tan A + cot A} = cos A \times sin A\) \(\boxed{\text{proved in 1 method}}\)

  3. \(sin^4A - cos^4A = 2sin^2A - 1\) \(\boxed{\text{proved in 1 method}}\)

  4. \((1 - tan A)^2 + (1 + tan A)^2 = 2sec^2A\) \(\boxed{\text{proved in 1 method}}\)

  5. \(cosec^4A - cosec^2A = cot^4A + cot^2A\) \(\boxed{\text{proved in 1 method}}\)

  6. \(sec^2A + cosec^2A = sec^2A \times cosec^2A\) \(\boxed{\text{proved in 1 method}}\)

  7. \(tan^2A - sin^2A = tan^2A \times sin^2A\) \(\boxed{\text{proved in 1 method}}\)

  8. \(cot^2A - cos^2A = cot^2A \times cos^2A\) \(\boxed{\text{proved in 1 method}}\)

  9. \((cosec A + sin A)(cosec A - sin A) = cot^2A + cos^2A\) \(\boxed{\text{proved in 2 methods}}\)

  10. \((sec A - cos A)(sec A + cos A) = sin^2A + tan^2A\) \(\boxed{\text{proved in 1 method}}\)


If anybody proved any one of these I will mark that question as proved. But in Trigonometry one can prove a proof in many methods. So I will also mention in how many methods the proof is proved. Try your best to prove these. These are very easy if you apply little brain.


For more of these see my set Proof Based Notes

Note by Ram Mohith
3 weeks, 4 days ago

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  1. \(sec^2A - cos^2A = 1 + tan^2A - (1 - sin^2A) = 1 - 1 + tan^2A + sin^2A\)
\(tan^2A + sin^2A\)

Hence Proved

Ram Mohith - 2 weeks, 6 days ago

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10) \((sec A-\cos A)(\sec A+\cos A)=\sec^2 A-\cos^2 A=(1+\tan^2 A)-(1-\sin^2 A)=\tan^2 A+\sin^2 A\)

X X - 2 weeks, 6 days ago

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Another way for 9) \((\csc A+\sin A)(\csc A-\sin A)=csc^2 A-sin^2 A=(\cot^2 A+1)-(1-\cos^2 A)=\cot^2 A+\cos^2 A \)

X X - 2 weeks, 6 days ago

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9) \((\csc A+\sin A)(\csc A-\sin A)=\csc^2 A-\sin^2 A\)

\(=\frac1{\sin^2 A}-\sin^2 A\)

\(=\frac{1-\sin^4 A}{\sin^2 A}\)

\(=\frac{(1-\sin^2 A)(1+\sin^2 A)}{\sin^2 A}\)

\(=\frac{\cos^2 A(1+\sin^2 A)}{\sin^2 A}\)

\(=\cot^2 A(1+\sin^2 A)\)

\(=\cot^2 A+\cot^2 A\times\sin^2 A\)

\(=\cot^2 A+\frac{\cos^2 A}{\sin^2 A}\times\sin^2 A\)

\(=\cot^2 A+\cos^2 A\)

X X - 2 weeks, 6 days ago

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8) \(\cot^2 A-\cos^2 A=\cos^2 A(\frac1{\sin^2 A}-1)=\cos^2 A\times\frac{1-\sin^2 A}{\sin^2 A}=\cos^2 A\times\frac{\cos^2 A}{\sin^2 A}=\cos^2 A\times\cot^2 A\)

X X - 2 weeks, 6 days ago

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7) \(\tan^2 A-\sin^2 A=\sin^2 A(\frac1{\cos^2 A}-1)=\sin^2 A\times\frac{1-\cos^2 A}{\cos^2 A}=\sin^2 A\times\frac{\sin^2 A}{\cos^2 A}=\sin^2 A\times\tan^2 A\)

X X - 2 weeks, 6 days ago

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6) \(\sec^2 A+\csc^2 A=\frac1{\cos^2 A}+\frac1{\sin^2 A}=\frac{\sin^2 A+\cos^2 A}{\sin^2 A\times\cos^2 A}=\frac1{\sin^2 A\times\cos^2 A}=\sec^2 A\times\csc^2 A\)

X X - 2 weeks, 6 days ago

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5) \(\csc^4 A-\csc^2 A=\csc^2 A(\csc^2 A-1)=(\cot^2 A+1)\times\cot^2 A=\cot^4 A+\cot^2 A\)

X X - 2 weeks, 6 days ago

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1) \(\frac{\csc A+1}{\csc A-1}=\frac{\frac1{\sin A}+1}{\frac1{\sin A}-1}=\frac{1+\sin A}{1-\sin A}\)

X X - 2 weeks, 6 days ago

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2)\(\frac1{\tan A+\cot A}=\frac1{\frac{\sin A}{\cos A}+\frac{\cos A}{\sin A}}=\frac{\cos A\times\sin A}{\sin^2 A+cos^2 A}=\cos A\times\sin A\)

X X - 2 weeks, 6 days ago

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4) \((1 - tan A)^2 + (1 + tan A)^2 = 2sec^2A\)

\(LHS \implies 1 + tan^2A - 2tan A + 1 + tan^2A + 2tan A\)

\(\implies sec^2A = sec^2A\) (since \(1 + tan^2A = sec^2A\)

\(\implies 2sec^2A\)

\(\color{red}\text{Hence Proved}\)

Ram Mohith - 3 weeks, 3 days ago

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3) \(sin^4A - cos^4A = 2sin^2A - 1\)

\(LHS \implies sin^4A - cos^4A \implies (sin^2A)^2 - (cos^2A)^2\)

\(\implies (sin^2A + cos^2A)(sin^2A - cos^2A)\)

\(\implies 1 \times (sin^2A - cos^2A) \implies sin^2A - (1 - sin^2A)\) (since \(cos^2A = 1 - sin^2A\))

\(\implies sin^2A - 1 + sin^2A\)

\(\implies 2sin^2A - 1\)

\(\color{green}\text{Hence Proved}\)

Ram Mohith - 3 weeks, 4 days ago

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