In a - PQR - PQ -12 cm - PR -13 cm- and - Q -90-Then- tan P - cot R is-A- 0B- 11-12C-D- 12-13

The correct option is A 0Given that in Δ PQR, PQ = 12 cm, PR = 13 cm, and ∠ Q = 90^∘, we have: tan P = QR/PQ and cot R = nbsp;QR/PQ Thus, tan P cot R = 0. ...

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Standard X

Mathematics

Trigonometric Identities

In a Δ PQR , ...

Question

In a $Δ P Q R, P Q = 12 c m,$ PR = 13 cm, and $∠ Q = 90^{\circ}$ .
Then, tan P - cot R is:

0.0

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\frac{11}{12}

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\frac{5}{6}

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\frac{12}{13}

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Solution

The correct option is A 0.0
Given that in $Δ P Q R, P Q = 12 c m, P R = 13 c m, a n d ∠ Q = 90^{\circ}$ , we have:
tan $P = \frac{Q R}{P Q} a n d c o t R = \frac{Q R}{P Q}$
Thus, tan P - cot R = 0.

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In a ΔPQR,PQ=12cm, PR = 13 cm, and ∠Q=90∘. Then, tan P - cot R is:

The correct option is A 0.0Given that in ΔPQR,PQ=12 cm,PR=13cm,and ∠Q=90∘, we have: tan P=QRPQ and cot R=QRPQ Thus, tan P - cot R = 0.

In a $Δ P Q R, P Q = 12 c m,$ PR = 13 cm, and $∠ Q = 90^{\circ}$ .
Then, tan P - cot R is:

The correct option is A 0.0
Given that in $Δ P Q R, P Q = 12 c m, P R = 13 c m, a n d ∠ Q = 90^{\circ}$ , we have:
tan $P = \frac{Q R}{P Q} a n d c o t R = \frac{Q R}{P Q}$
Thus, tan P - cot R = 0.