Question

From the following figure,find the values of:
(i) cos A
(ii) cosec A
(iii) tan^2 A- sec^2 A
(iv) sin C
(v) sec C
(vi) cot^2 C - 1/sin^2 C​​

Answer 1

Answer:

Analysis→

Here we're given a figure formed by joining two triangles or a smaller triangle inscribed in a larger triangle. With points A, B, C, so add a point D. Hence we've two distinct triangles ABD and BDC. And we've to find the following trigonometric ratios:-

• CosA
• CosecA
• Tan²A-Sec²A
• SinC
• SecC
• Cot²C-1/Sin²C
• And 1/Sin=Cosec; so 1/Sin²=Cosec²

Given→

• AB=5
• AD=3
• BD=4
• BC=15

To Find→

• CosA
• CosecA
• Tan²A-Sec²A
• SinC
• SecC
• Cot²C-1/Sin²C

Answer→

1. CosA:-

$\rm\rightarrow{\cos\theta=\dfrac{B}{H}}$

$\rm\rightarrow{\cos A=\dfrac{AD}{AB}}$

$\rm\rightarrow{\cos A=\dfrac{3}{5}}$

${\boxed{\boxed{\rightarrow{\rm{\cos A=\dfrac{3}{5}\checkmark}}}}}$

2. CosecA:-

$\rm\rightarrow{\cosec\theta=\dfrac{H}{P}}$

$\rm\rightarrow{\cosec A=\dfrac{AB}{BD}}$

$\rm\rightarrow{\cosec A=\dfrac{5}{4}}$

${\boxed{\boxed{\rightarrow{\rm{\cosec A=\dfrac{5}{4}\checkmark}}}}}$

3. Tan²A-Sec²A:-

$\rm\rightarrow{\tan^2\theta-\sec^2\theta=(\frac{P}{B})^2-(\frac{H}{B})^2}$

$\rm\rightarrow{\tan^2\theta-\sec^2\theta=(\frac{BD}{AD})^2-(\frac{AB}{AD})^2}$

$\rm\rightarrow{\tan^2A-\sec^2A=(\frac{4}{3})^2-(\frac{5}{3})^2}$

$\rm\rightarrow{\tan^2A-\sec^2A=\dfrac{16}{9}-\dfrac{25}{9}}$

$\rm\rightarrow{\tan^2A-\sec^2A=\dfrac{16-25}{9}}$

$\rm\rightarrow{\tan^2A-\sec^2A=\dfrac{-9}{9}}$

$\rm\rightarrow{\tan^2A-\sec^2A=\dfrac{\cancel{-9}}{\cancel{9}}}$

$\rm\rightarrow{\tan^2A-\sec^2A=-1}$

${\boxed{\boxed{\rightarrow{\rm{\tan^2A-\sec^2A=-1\checkmark}}}}}$

4. SinC:-

$\rm\rightarrow{\sin\theta=\dfrac{P}{H}}$

$\rm\rightarrow{\sin A=\dfrac{BD}{BC}}$

$\rm\rightarrow{\sin A=\dfrac{4}{12}}$

$\rm\rightarrow{\sin A=\dfrac{\cancel{4}}{\cancel{12}}}$

$\rm\rightarrow{\sin A=\dfrac{1}{3}}$

${\boxed{\boxed{\rightarrow{\rm{\sin A=\dfrac{1}{3}\checkmark}}}}}$

5. SecC

$\rm\rightarrow{\sec\theta=\dfrac{H}{B}}$

$\rm\rightarrow{\sec C=\dfrac{BC}{DC}}$

$\rm\rightarrow{\sec C=\dfrac{12}{\sqrt{128}}}$

${\boxed{\boxed{\rightarrow{\rm{\sec C=\dfrac{12}{\sqrt{128}}\checkmark}}}}}$

6. Cot2C-Cosec²C

$\rm\rightarrow{\cot^2\theta-\cosec^2\theta=(\frac{B}{P})^2-(\frac{H}{P})^2}$

$\rm\rightarrow{\cot^2C-\cosec^2C=(\frac{DC}{BD})^2-(\frac{BC}{BD})^2}$

$\rm\rightarrow{\cot^2C-\cosec^2C=(\frac{\sqrt{128}}{4})^2-(\frac{12}{4})^2}$

$\rm\rightarrow{\cot^2C-\cosec^2C=\dfrac{128}{16}-\dfrac{144}{16}}$

$\rm\rightarrow{\cot^2C-\cosec^2C=\dfrac{128-144}{16}}$

$\rm\rightarrow{\cot^2C-\cosec^2C=\dfrac{-16}{16}}$

$\rm\rightarrow{\cot^2C-\cosec^2C=\dfrac{\cancel{-16}}{\cancel{16}}}$

$\rm\rightarrow{\cot^2C-\cosec^2C=-1}$

${\boxed{\boxed{\rightarrow{\rm{\cot^2C-\cosec^2C=-1\checkmark}}}}}$

HOPE IT HELPS.