Question

Sec 15×cosec75 -cot 75 ×tan 15

Answer 1

${ \tt{ \large { \purple{Solution}}}}$

${ \rm{ \sec(15) \times \csc(75) - \cot(75) \times \tan(15) }}$

${ \rm { = \sec( {90}^{ \degree} - {75}^{ \degree} ) \times \csc( {75}^{ \degree} ) - \cot( {75}^{ \degree} ) \times \tan( {90}^{ \degree} - {15}^{ \degree} ) }}$

${ \rm{ = \csc(75) \times \csc(75) - \cot(75) \times \cot(75) }}$

${ \rm{ = { \csc}^{2} {75}^{ \degree} - { \cot }^{2} {75}^{ \degree} }}$

${ \rm{ = 1}}$

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${ \tt{ \large{ \blue{more \: related \: formulas}}}}$

${ \rm{(i) { \sin }^{2} a + { \cos}^{2} a = 1}}$

${ \rm{(ii) { \sec }^{2} a - { \tan }^{2} a = 1}}$

${ \rm{(iii) { \csc }^{2} a - { \cot}^{2} a = 1}}$

${ \rm{ \sin = \frac{p}{h} }}$

${ \rm{ \cos = \frac{b}{h} }}$

${ \rm{ \tan = \frac{h}{b} }}$

${ \rm{ \sec = \frac{p}{b} }}$

${ \rm{ \cot = \frac{b}{h} }}$

${ \rm{h = height}}$

${ \rm{ b= base}}$

${ \rm{p = perpendicular}}$