Prove that-tan 69-tan 66-1-tan 69-tan 66-1

LHS: tan69^∘+tan66^∘/1 tan69^∘tan66^∘ = tan(69^∘+66^∘) =tan(135^∘) = tan(90^∘+45^∘) = cot45^∘ [∵tan θ is negative in second quadrant] = 1 =RHS therefore LHS=RH ...

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Standard XII

Mathematics

Sign of Trigonometric Ratios in Different Quadrants

Prove that:ta...

Question

Prove that:
$\frac{t a n 69^{\circ} + t a n 66^{\circ}}{1 - t a n 69^{\circ} t a n 66^{\circ}} = - 1$

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Solution

$L H S : \frac{t a n 69^{\circ} + t a n 66^{\circ}}{1 - t a n 69^{\circ} t a n 66^{\circ}}$
$= t a n (69^{\circ} + 66^{\circ})$
$= t a n (135^{\circ})$
$= t a n (90^{\circ} + 45^{\circ})$
$= - c o t 45^{\circ}$
$[∵ t a n θ is negative in second quadrant]$
=-1
=RHS
$t h e r e f o r e$ LHS=RHS
Hence proved.

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Similar questions

Q. Prove that

\frac{\tan 69 ° + \tan 66 °}{1 - \tan 69 ° \tan 66 °} = - 1

State true or false.

tan 69^{\circ} + tan 66^{\circ} + 1 = tan 69^{\circ} tan 66^{\circ}

Q. If

tan 69^{\circ} + tan 66^{\circ} - tan 69^{\circ} tan 66^{\circ} = 2 k, t h e n 2 k =

Q. If

t a n 69^{\circ} + t a n 66^{\circ} - t a n 69^{\circ} t a n 66^{\circ} = 2 k,

then k=

Q. If tan 69° + tan 66° − tan 69° tan 66° = 2k, then k =

(a) −1
(b)

\frac{1}{2}

(c)

- \frac{1}{2}

(d) None of these

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Prove that: tan69∘+tan66∘1−tan69∘tan66∘=−1

LHS:tan69∘+tan66∘1−tan69∘tan66∘ =tan(69∘+66∘) =tan(135∘) =tan(90∘+45∘) =−cot45∘ [∵ tanθ is negative in second quadrant] =-1 =RHS therefore LHS=RHS Hence proved.

Prove that:
$\frac{t a n 69^{\circ} + t a n 66^{\circ}}{1 - t a n 69^{\circ} t a n 66^{\circ}} = - 1$