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Exercise 3.3 · Q3

Q.Find the values of other five trigonometric functions for the following:

(i) cos⁡θ=−12\cos\theta=-\dfrac12, θ\theta lies in the III quadrant.
(ii) cos⁡θ=23\cos\theta=\dfrac23, θ\theta lies in the I quadrant.
(iii) sin⁡θ=−23\sin\theta=-\dfrac23, θ\theta lies in the IV quadrant.
(iv) tan⁡θ=−2\tan\theta=-2, θ\theta lies in the II quadrant.
(v) sec⁡θ=135\sec\theta=\dfrac{13}{5}, θ\theta lies in the IV quadrant.
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Each part gives one function's value and the quadrant; a Pythagorean identity gives the paired basic ratio up to a sign, and the ASTC table (§3.4.1) fixes that sign for the stated quadrant, after which the remaining functions follow from the quotient/reciprocal identities.

Step 1. Part (i): cos⁡θ=−12\cos\theta=-\dfrac12, III quadrant.

sin⁡2θ=1−cos⁡2θ=1−14=34⇒sin⁡θ=±32\sin^2\theta=1-\cos^2\theta=1-\dfrac14=\dfrac34\Rightarrow\sin\theta=\pm\dfrac{\sqrt3}{2}. Sine is negative in Quadrant III, so sin⁡θ=−32\sin\theta=-\dfrac{\sqrt3}{2}.

tan⁡θ=sin⁡θcos⁡θ=−3/2−1/2=3\tan\theta=\dfrac{\sin\theta}{\cos\theta}=\dfrac{-\sqrt3/2}{-1/2}=\sqrt3; cosec⁡θ=1sin⁡θ=−23=−233\operatorname{cosec}\theta=\dfrac{1}{\sin\theta}=-\dfrac{2}{\sqrt3}=-\dfrac{2\sqrt3}{3}; sec⁡θ=1cos⁡θ=−2\sec\theta=\dfrac{1}{\cos\theta}=-2; cot⁡θ=1tan⁡θ=13=33\cot\theta=\dfrac{1}{\tan\theta}=\dfrac{1}{\sqrt3}=\dfrac{\sqrt3}{3}.

Step 2. Part (ii): cos⁡θ=23\cos\theta=\dfrac23, I quadrant.

sin⁡2θ=1−49=59⇒sin⁡θ=±53\sin^2\theta=1-\dfrac49=\dfrac59\Rightarrow\sin\theta=\pm\dfrac{\sqrt5}{3}; every function is positive in Quadrant I, so sin⁡θ=53\sin\theta=\dfrac{\sqrt5}{3}.

tan⁡θ=5/32/3=52\tan\theta=\dfrac{\sqrt5/3}{2/3}=\dfrac{\sqrt5}{2}; cosec⁡θ=35=355\operatorname{cosec}\theta=\dfrac{3}{\sqrt5}=\dfrac{3\sqrt5}{5}; sec⁡θ=32\sec\theta=\dfrac32; cot⁡θ=25=255\cot\theta=\dfrac{2}{\sqrt5}=\dfrac{2\sqrt5}{5}.

Step 3. Part (iii): sin⁡θ=−23\sin\theta=-\dfrac23, IV quadrant.

cos⁡2θ=1−49=59⇒cos⁡θ=±53\cos^2\theta=1-\dfrac49=\dfrac59\Rightarrow\cos\theta=\pm\dfrac{\sqrt5}{3}; cosine is positive in Quadrant IV, so cos⁡θ=53\cos\theta=\dfrac{\sqrt5}{3}.

tan⁡θ=−2/35/3=−25=−255\tan\theta=\dfrac{-2/3}{\sqrt5/3}=-\dfrac{2}{\sqrt5}=-\dfrac{2\sqrt5}{5}; cosec⁡θ=1−2/3=−32\operatorname{cosec}\theta=\dfrac{1}{-2/3}=-\dfrac32; sec⁡θ=35=355\sec\theta=\dfrac{3}{\sqrt5}=\dfrac{3\sqrt5}{5}; cot⁡θ=5/3−2/3=−52\cot\theta=\dfrac{\sqrt5/3}{-2/3}=-\dfrac{\sqrt5}{2}.

Step 4. Part (iv): tan⁡θ=−2\tan\theta=-2, II quadrant.

sec⁡2θ=1+tan⁡2θ=1+4=5⇒sec⁡θ=±5\sec^2\theta=1+\tan^2\theta=1+4=5\Rightarrow\sec\theta=\pm\sqrt5; cosine (and secant) is negative in Quadrant II, so sec⁡θ=−5\sec\theta=-\sqrt5, cos⁡θ=−15=−55\cos\theta=-\dfrac{1}{\sqrt5}=-\dfrac{\sqrt5}{5}.

sin⁡θ=tan⁡θ⋅cos⁡θ=(−2)(−15)=25=255\sin\theta=\tan\theta\cdot\cos\theta=(-2)\left(-\dfrac{1}{\sqrt5}\right)=\dfrac{2}{\sqrt5}=\dfrac{2\sqrt5}{5} (positive, consistent with Quadrant II).

cosec⁡θ=1sin⁡θ=52\operatorname{cosec}\theta=\dfrac{1}{\sin\theta}=\dfrac{\sqrt5}{2}; cot⁡θ=1tan⁡θ=−12\cot\theta=\dfrac{1}{\tan\theta}=-\dfrac12.

Step 5. Part (v): sec⁡θ=135\sec\theta=\dfrac{13}{5}, IV quadrant. …

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